Lesson 4 Thermal Energy
| Site: | MoodleHUB.ca 🍁 |
| Course: | Science 10 [5 cr] - AB Ed copy 1 |
| Book: | Lesson 4 Thermal Energy |
| Printed by: | Guest user |
| Date: | Sunday, 7 September 2025, 6:44 PM |
Introduction
We have learned how the earth’s and sun’s positions affect climate, but how does the movement of thermal energy on Earth affect it?
DS2.1 Mountain range seen from space with clouds on one side
In this section, you will learn more details about thermal energy and how it is absorbed. Does every substance absorb thermal energy the same way? How can we tell how much thermal energy is absorbed?
Once you have a better understanding of thermal energy, you will look at how thermal energy moves around the globe and contributes to different climates. Why does living near an ocean, large lake, or mountains affect your climate? How do weather patterns across Earth affect us?
Finally, you will learn what a climatograph is and how to read one. You will use climatographs to compare biomes.
Once you have a better understanding of thermal energy, you will look at how thermal energy moves around the globe and contributes to different climates. Why does living near an ocean, large lake, or mountains affect your climate? How do weather patterns across Earth affect us?
Finally, you will learn what a climatograph is and how to read one. You will use climatographs to compare biomes.
Targets
By the end of this section, you will be able to
DS2.2 Heating curve of water
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investigate and calculate the different thermal properties of substances on Earth and how this affects climate; this includes the different rates at which thermal energy transfers from one substance to another
DS2.3 Water cycle
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describe how thermal energy is transferred through the atmosphere and the hydrosphere and use this information to describe the climates of biomes
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analyze climatographs of biomes and explain why biomes with similar characteristics can exist in different locations on Earth
Introduction
How does thermal energy work?
© By Flanker, penubag (Own work), via Wikimedia Commons
D4.1 Phase changes
D4.1 Phase changes
In this lesson, we will study thermal energy in more detail. Remember, thermal energy is the same as the kinetic energy of the particles in a substance. The more movement a particle has (kinetic energy), the warmer the substance is and the more thermal energy it has.
This lesson will take your knowledge one step further and talk about how different substances absorb and emit thermal energy at different rates. It will discuss the effects of this uneven absorption and teach you the calculations so you can determine yourself how a substance absorbs thermal energy.
Finally, this lesson will look at the different phase changes and how those processes transfer thermal energy through the water cycle.
This lesson will take your knowledge one step further and talk about how different substances absorb and emit thermal energy at different rates. It will discuss the effects of this uneven absorption and teach you the calculations so you can determine yourself how a substance absorbs thermal energy.
Finally, this lesson will look at the different phase changes and how those processes transfer thermal energy through the water cycle.
Targets
By the end of this lesson, you will be able to- describe how differences in the absorption of thermal energy by different substances can lead to uneven heating and cooling
- explain how different phase changes, specifically in water, transfer thermal energy
- complete calculations to visualize this transfer using «math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «mo»,«/mo» «mo»§#160;«/mo» «msub» «mi»H«/mi» «mrow» «mi»v«/mi» «mi»a«/mi» «mi»p«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «/math», and Q = mcΔt
Watch This
Heat Exchange © YouTube Bozeman Science
This video will give you a great review of how thermal energy is transferred and will introduce you to the idea of a substance’s specific heat capacity, which we will learn more about in this lesson. Please note that a temperature unit called “Kelvins” is used in this video rather than degrees Celsius.
Thermal Properties
The differences in how substances absorb thermal energy can have a large impact on climate.
| Compound |
Specific Heat Capacity
(J/g•°C) |
|---|---|
| water | 4.19 |
| methanol | 2.56 |
| air | 1.01 |
| aluminium | 0.897 |
| soil | 0.85 |
| stainless steel
|
0.502 |
Every substance has different thermal properties. The thermal property we will focus on in this course is how substances absorb thermal energy differently. Some substances can absorb lots of thermal energy very quickly, while others absorb it more slowly. These same substances will then release the thermal energy just as fast or as slow as they absorbed it. This property is called a substance’s specific heat capacity, and it is the amount of energy needed to raise 1 g of a substance 1 ˚C. Remember, thermal energy always moves from an area of high thermal energy to an area of low thermal energy.
Water has a specific heat capacity of 4.19 J/g•˚C. This means it takes 4.19 J of energy to raise 1 g (or 1 mL) of water 1 ˚C. Water has a relatively high specific heat capacity, while aluminium has a lower specific heat capacity, just 0.897 J/g•˚C. This means it takes only 0.897 J of energy to raise 1 g of aluminium 1 ˚C. Generally speaking, the hydrosphere has a much higher specific heat capacity than the lithosphere does. This is why cities that are close to large bodies of water have a different climate than cities surrounded by land, even when they are in a similar biome.
Water has a specific heat capacity of 4.19 J/g•˚C. This means it takes 4.19 J of energy to raise 1 g (or 1 mL) of water 1 ˚C. Water has a relatively high specific heat capacity, while aluminium has a lower specific heat capacity, just 0.897 J/g•˚C. This means it takes only 0.897 J of energy to raise 1 g of aluminium 1 ˚C. Generally speaking, the hydrosphere has a much higher specific heat capacity than the lithosphere does. This is why cities that are close to large bodies of water have a different climate than cities surrounded by land, even when they are in a similar biome.
Since water has such a high specific heat capacity, it takes a lot of energy to raise the temperature of water. This means water takes longer to heat up and cool down compared to other substances. When water does cool down, a lot of energy is released into the surrounding environment. Because it takes so long for water to heat up or cool down, it actually helps to level out any variations in temperature for cities located nearby. If the air temperature is very cold, the thermal energy will leave the water and heat up the air and land. If the air temperature is very warm, the water will absorb the thermal energy and cool down the air and land. The water will do this with very little change to its own temperature. If the water itself is cooler (such as the Atlantic Ocean), it will create a cooler climate, as it will absorb more thermal energy. If the water is warmer (such as the Pacific Ocean), it will create a warmer climate, as it will release more thermal energy.
Water’s specific heat capacity plays a very large part in the climate of the areas surrounding large bodies of water.
Water’s specific heat capacity plays a very large part in the climate of the areas surrounding large bodies of water.
D4.3 Inner Harbour at Victoria BC
Did You Know?
D4.4 A greenhouse in the winter
Large containers of water are sometimes placed in greenhouses over the winter to help prevent the plants from freezing. During the day, the sun heats the greenhouse and the water absorbs the excess heat. At night, the water cools and releases large amounts of thermal energy into the air, warming the greenhouse up.
Read This
Please read pages 377 and 378 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on what specific heat capacity is and how it affects climate. Remember, if you have any questions or you do not understand something, ask your teacher!
Practice Questions
Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.
- Explain what specific heat capacity is.
Specific heat capacity is the amount of energy it takes to raise 1 g of a substance 1 ˚C.
-
How does specific heat capacity affect climate?
Water has a relatively high specific heat capacity, so it takes lots of thermal energy for water to heat up or cool down. This means it can transfer or absorb lots of thermal energy from the air and land around it without changing its own temperature very much. This makes the land and air around water have less variation in temperature. It also means cooler water makes cooler climates, while warmer water makes warmer climates.
Virtual Lab
Heat Transfer by Conduction © Explore Learning
Background Information:
This lab will help you visualize how thermal energy moves between substances. It will also help you understand how the specific heat capacity of a substance affects how it heats and cools as well as how it transfers heat.
Please note: if you scroll down while in the Gizmo you will see a list of questions. You DO NOT need to complete these questions. You are able to complete them for extra practice if you would like.
This lab will help you visualize how thermal energy moves between substances. It will also help you understand how the specific heat capacity of a substance affects how it heats and cools as well as how it transfers heat.
Please note: if you scroll down while in the Gizmo you will see a list of questions. You DO NOT need to complete these questions. You are able to complete them for extra practice if you would like.
- Click on the play icon to open the Gizmo. Print students can access the Gizmo in the Online Resources for Print Student section of their online course.
- To set up the simulation, click on the “GRAPH” tab.
© Explore learning
D4.5 The graph tab
D4.5 The graph tab
- Make sure the substance selected is “Aluminum.” Please note that the simulation spells aluminium as aluminum.
- Note the initial temperature of beaker A and beaker B and record it in a chart similar to the one below.
Initial Temperature (°C)
Final Temperature (°C)
Time It Took for Temperatures
to Reach Equilibrium
aluminium beaker A
beaker B
copper beaker A
beaker B
Steel beaker A
beaker B
Glass beaker A
beaker B
- Click the play button in the simulation and watch what happens to the temperature of both beakers.
- Watch for when the two temperatures become the same (reach equilibrium) and take note of how long it took for that to happen. Record this observation and the final temperature of each beaker in your observation chart.
- Look at the graph that has been created and note what has happened to the temperatures of each beaker. You can zoom out on the graph to see all of the data recorded if needed.
- Click reset and change the substance selected to “Copper.”
- Repeat steps 4 to 7 for copper.
- Click reset and change the substance selected to “Steel.”
- Repeat steps 4 to 7 for steel.
- Click reset and change the substance selected to “Glass.”
- Repeat steps 4 to 7 for glass.
- Please return to the top of this page and click on analysis to complete the analysis questions.
© Explore learning
D4.6 Aluminum is selected
D4.6 Aluminum is selected
Here is an example of one row of the observation chart filled out.
|
|
|
Initial Temperature (°C)
|
Final Temperature (°C)
|
Time It Took for Temperatures
to Reach Equilibrium |
|---|---|---|---|---|
| aluminium | beaker A
|
95
|
50
|
22 min 50 s
|
|
beaker B
|
5
|
50
|
||
| copper | beaker A |
|
|
|
| beaker B
|
|
|
||
| Steel | beaker A |
|
|
|
| beaker B
|
|
|
||
| Glass | beaker A |
|
|
|
| beaker B
|
|
|
- Which substance had the highest specific heat capacity? Which one has the lowest?
Glass has the highest specific heat capacity, and copper had the lowest. - How did you know which substance had the highest specific heat capacity and which one had the lowest?
The substance that caused the beakers to take the longest to reach the same temperature has the highest specific heat capacity. This is because it takes a lot of energy for that substance to heat up, and it takes a long time for that substance to release the energy when cooling down.
The substance that caused the beakers to reach the same temperature the fastest has the lowest specific heat capacity. This is because that substance will gain and release thermal energy much faster.
- Which direction did heat flow in?
The thermal energy, or heat, flowed from the hot beaker to the cold beaker. - When did the heat stop flowing? Why did it stop?
The heat stopped flowing once both beakers were the same temperature. This is because there is no excess thermal energy to transfer to an area with less thermal energy.
- What happened to the speed at which the temperature change occurs as the temperatures got closer to equilibrium? Why does this happen?
The speed of the temperature change slowed down as the beakers got closer to being the same temperature. This is because there is less excess thermal energy to transfer.
Calculating the Amount of Thermal Energy
Does the amount of energy needed to heat up a substance increase with the substance’s mass?
D4.7 How much thermal energy is in the ocean?
The specific heat capacity tells us how much thermal energy is released or absorbed when we change the temperature of 1 g of a substance. But an ocean is much more than 1 g of water. How can we tell how much energy is needed to change the temperature
of a larger body of water?
The answer is the specific heat capacity formula.
The answer is the specific heat capacity formula.
Q = mc∆t
Q = the quantity of thermal energy in joules (J)
m = the mass of the substance in grams (g)
c = the specific heat capacity of the substance in J/g•˚C
Δt = change in temperature in ˚C
Q = the quantity of thermal energy in joules (J)
m = the mass of the substance in grams (g)
c = the specific heat capacity of the substance in J/g•˚C
Δt = change in temperature in ˚C
We can use this formula to determine
- the amount of thermal energy it takes to raise the temperature of a certain mass of a substance
- the change in temperature of a substance
- the specific heat capacity of a substance
- the mass of a substance
Let’s look at some examples. Each example has a video to go with it. To play the video, click on the play icon next to the example.
Examples
- How much thermal energy is needed to raise the temperature of North Buck Lake, located in Athabasca County, from 17.0 ˚C to 19.0 ˚C. The lake contains 1.57 × 1014 g of water.
https://adlc.wistia.com/medias/at3nilai60
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo».«/mo»«mn»57«/mn»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«msup»«mn»10«/mn»«mn»14«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»19«/mn»«mo».«/mo»«mn»0«/mn»«mo»-«/mo»«mn»17«/mn»«mo».«/mo»«mn»0«/mn»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
Note: To find the change in temperature, we have to find the difference between the starting temperature and the final temperature.
Step 2: Substitute values into the formula.
Q = mcΔt
Q = (1.57 x 1014 g)(4.19 J/g•˚C)(2.0 ˚C)
Step 3: Calculate the answer.
Q = (1.57 x 1014 g)(4.19 J/g•˚C)(2.0 ˚C)
Q = 1.32 x 1015 J to three significant digits
It will take 1.32 x 1015 J, or 1 320 000 000 000 000 (1.32 quadrillion) J, of thermal energy to raise North Buck Lake 2 ˚C.
- A glass of 250 g of water is placed in the fridge to cool off. If the temperature of the water drops by 5.0 ˚C, how much thermal energy is released?
https://adlc.wistia.com/medias/psy0dpk2rb
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»250«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Substitute values into the formula.
Q = mcΔt
Q = (250 g)(4.19 J/g•˚C)(5.0 ˚C)
Step 3: Calculate the answer.
Q = (250 g)(4.19 J/g•˚C)(5.0 ˚C)
Q = 5 237.5 J
We must round this answer to two significant digits, because two digits is the smallest number of digits found in the question.
To round this answer to two significant digits, we need to convert it to scientific notation.
5 237.5 = 5.2 × 103 J
The glass of water will release 5.2 x 103 J of thermal energy as it cools 5.0 ˚C.
For a review on significant digits and converting to scientific notation, go to Unit C, Section 1, Lesson 2.
- If 962 J of thermal energy is used to raise the temperature of some water from 6.0 ˚C to 10 ˚C, how much water was there?
https://adlc.wistia.com/medias/xtjnm7ci8n
Step 1: List your variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mo»§#160;«/mo»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»962«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»10«/mn»«mo»-«/mo»«mn»6«/mn»«mo».«/mo»«mn»0«/mn»«mo»=«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
Note: To find the change in temperature, we have to find the difference between the starting temperature and the final temperature.
Step 2: Rearrange the formula.
Q = mcΔt
We need to isolate m. To do this, we need to move c and Δt to the other side of the formula. Since they are being multiplied, we use division to move them to the other side.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mfrac» «mi»Q«/mi» «mrow» «mi»c«/mi» «mo»§#8710;«/mo» «mi»t«/mi» «/mrow» «/mfrac» «mo»=«/mo» «mi»m«/mi» «/math»
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses!
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«mrow»«mi»c«/mi»«mo»§#8710;«/mo»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»962«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«mstyle displaystyle=¨true¨»«mfenced»«mrow»«mn»4«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mstyle»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»962«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»4«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»962«/mn»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»16«/mn»«mo».«/mo»«mn»76«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»57«/mn»«mo».«/mo»«mn»399«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«/mtable»«/math»
The mass needs to be rounded to two significant digits, as two digits is the smallest number of digits found in the question.
m = 57 g
There was 57 g of water.
- A 1.326 kg stainless steel pot was placed on a stove to get warm. If the pot started out at 20.0 ˚C and 1 000 J of thermal energy was added to the pot, what is its final temperature? The specific heat capacity of stainless steel is 0.502
J/g•˚C https://adlc.wistia.com/medias/8lhg5l8ztr
D4.9 Stainless steel pot on stove
Step 1: List the variables.
Q = 1 000 J
m = 1.326 kg = 1 326 g
Note: The mass must be converted to g to be used in the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mo»?«/mo»«mrow»«mn»1«/mn»«mo».«/mo»«mn»326«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Now cross-multiply and divide.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mn»1«/mn» «mo»§#160;«/mo» «mn»000«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»g«/mi» «mo»§#215;«/mo» «mn»1«/mn» «mo».«/mo» «mn»326«/mn» «mo»§#160;«/mo» «mi»kg«/mi» «mo»=«/mo» «mn»1«/mn» «mo»§#160;«/mo» «mn»326«/mn» «mspace linebreak=¨newline¨»«/mspace» «mfrac» «mrow» «mn»1«/mn» «mo»§#160;«/mo» «mn»326«/mn» «/mrow» «mn»1«/mn» «/mfrac» «mo»=«/mo» «mn»1«/mn» «mo»§#160;«/mo» «mn»326«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»g«/mi» «/math»
Please click here for a refresher on how to do unit conversions.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»502«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Rearrange the formula.
Q = mcΔt
We need to isolate Δt. To do this, we need to move m and c to the other side of the formula. Since they are being multiplied, we use division to move them to the other side.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mfrac» «mi»Q«/mi» «mrow» «mi»m«/mi» «mi»c«/mi» «/mrow» «/mfrac» «mo»=«/mo» «mo»§#8710;«/mo» «mi»t«/mi» «/math»
Step 3: Substitute in the values.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«mrow»«mi»m«/mi»«mi»c«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»326«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mstyle displaystyle=¨true¨»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»502«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mstyle»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»326«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»502«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»665«/mn»«mo».«/mo»«mn»652«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo».«/mo»«mn»502«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
This gives us how much the temperature of the pot changed by. It does not give us the final temperature of the pot, which is what the question is asking for. To find the final temperature of the pot, you need to add the temperature change to the starting temperature.
20.0 ˚C + 1.502 ˚C = 21.502 ˚C
Now we need to round our answer to three significant digits.
The final temperature of the pot is 21.5 ˚C.
Step 1: Expand the formula.
There is another way you can calculate the final temperature of the pot using just one formula. To do this, you need to remember Δ t = tf – ti (Change in Temperature = Final Temperature – Initial Temperature).
Knowing this, we can substitute it into our formula.
Q = mc(tf – ti)Step 2: Rearrange the formula.
Then we would rearrange to find tf. We want to isolate tf, so first we need to move m and c to the other side. Just like before, they are being multiplied, so we use division to move them over.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mfrac» «mi»Q«/mi» «mrow» «mi»m«/mi» «mi»c«/mi» «/mrow» «/mfrac» «mo»=«/mo» «msub» «mi»t«/mi» «mi»f«/mi» «/msub» «mo»-«/mo» «msub» «mi»t«/mi» «mi»i«/mi» «/msub» «/math»
Now we need to move ti over. Since it is being subtracted, we use addition.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mfrac» «mi»Q«/mi» «mrow» «mi»m«/mi» «mi»c«/mi» «/mrow» «/mfrac» «mo»+«/mo» «msub» «mi»t«/mi» «mi»i«/mi» «/msub» «mo»=«/mo» «msub» «mi»t«/mi» «mi»f«/mi» «/msub» «/math»
Step 3: Substitute in the values and calculate answer.
Now you can substitute in your values and calculate your answer.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»326«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»502«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mrow»«/mfrac»«mo»+«/mo»«mn»20«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msub»«mi»t«/mi»«mi»f«/mi»«/msub»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»t«/mi»«mi»f«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»21«/mn»«mo».«/mo»«mn»502«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
- A piece of copper was tested to determine its specific heat capacity. It was determined that the 50.0 g piece of copper released 96.25 J of thermal energy when its temperature changed 5.00 ˚C. What is the specific heat capacity of copper?
https://adlc.wistia.com/medias/bk9s1317y3
Step 1: List your variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»96«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»50«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Rearrange the formula.
Q = mcΔt
We need to isolate c. To do this, we need to move m and Δt to the other side of the formula. Since they are being multiplied, we use division to move them to the other side.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mfrac» «mi»Q«/mi» «mrow» «mi»m«/mi» «mo»§#8710;«/mo» «mi»t«/mi» «/mrow» «/mfrac» «mo»=«/mo» «mi»c«/mi» «/math»
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses!
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«mrow»«mi»m«/mi»«mo»§#8710;«/mo»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»c«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»96«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»50«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mstyle displaystyle=¨true¨»«mfenced»«mrow»«mn»5«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mstyle»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»c«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»96«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»50«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mstyle displaystyle=¨true¨»«mfenced»«mrow»«mn»5«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mstyle»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»c«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»96«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»250«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»c«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»385«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
The specific heat capacity of copper is 0.385 J/g•˚C.
Did You Know?
D4.8 Copper pot
Copper is often used along the bottom of pots and pans or to make pots and pans. This is because of its ability to heat quickly and evenly, in part due to its low specific heat capacity.
Read This
Please read pages 378 to 380 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the different calculations that
can be done with Q = mcΔt. Please also try the practice questions found on these pages. Remember, if you have any questions or you do not understand something, ask your teacher!
Practice Questions
Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to
the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.- A bottle filled with 350 g of water is warmed from 17.0 ˚C to 21.0 ˚C in a hot car. How much thermal energy has the water absorbed?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»350«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»21«/mn»«mo».«/mo»«mn»0«/mn»«mo»-«/mo»«mn»17«/mn»«mo».«/mo»«mn»0«/mn»«mo»=«/mo»«mn»4«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
Note: To find the change in temperature, we have to find the difference between the starting temperature and the final temperature.Step 2: Substitute values into the formula.
Q = mc∆t
Q = (350 g)(4.19 J/g•˚C)(4.00 ˚C)Step 3: Calculate the answer.
Q = (350 g)(4.19 J/g•˚C)(4.00 ˚C)
Q = 5 866 J
We must round this answer to three significant digits, because three digits is the smallest number of digits found in the question.
To round this answer to three significant digits, we need to convert it to scientific notation.
5 866 = 5.87 x 103 J
The bottle of water will absorb 5.87 x 103 J of thermal energy.
For a review on significant digits and converting to scientific notation, please go to Unit C, Section 1, Lesson 2. - A man wearing a tungsten wedding ring submerges his hand in ice water to soothe a burn. If 15 J of thermal energy is released by the 20 g ring, what was the change in temperature? Tungsten has a specific heat capacity of 0.134 J/g•˚C.
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»15«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»134«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»Step 2: Rearrange the formula.
Q = mcΔt
We need to isolate Δt. To do this, we need to move m and c to the other side of the formula. Since they are being multiplied, we use division to move them to the other side.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mi»Q«/mi»«mrow»«mi»m«/mi»«mi»c«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mo»§#8710;«/mo»«mi»t«/mi»«/math»Step 3: Substitute in the values.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«mrow»«mi»m«/mi»«mi»c«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»15«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mstyle displaystyle=¨true¨»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»134«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mstyle»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 4: Calculate the answer.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»15«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»134«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»15«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»2«/mn»«mo».«/mo»«mn»68«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»597«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
We must round this answer to two significant digits, because two digits is the smallest number of digits found in the question.
Δt = 5.6 ˚C
The ring drops 5.6 ˚C. - Slave Lake City is located on the shores of Lesser Slave Lake. This is a huge lake that can release lots of thermal energy. On a cold winter day, the lake releases 5.74 × 1013 kJ of thermal energy into the air, warming the town up a bit.
This causes the lake to change 1.0 ˚C in temperature. What is the mass of the water in the lake?
Step 1: List your variables.
Q = 5.74 x 1013 kJ = 5.74 x 1016 J
Note: The quantity of thermal energy must be converted to J to be used in the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mo»?«/mo»«mrow»«mn»5«/mn»«mo».«/mo»«mn»74«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»13«/mn»«/msup»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Now cross-multiply and divide.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»§#215;«/mo»«mn»5«/mn»«mo».«/mo»«mn»74«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»13«/mn»«/msup»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»74«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»16«/mn»«/msup»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mn»5«/mn»«mo».«/mo»«mn»74«/mn»«mo»§#160;«/mo»«mo»§#215;«/mo»«mo»§#160;«/mo»«msup»«mn»10«/mn»«mn»16«/mn»«/msup»«mo»/«/mo»«mn»1«/mn»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»74«/mn»«mo»§#160;«/mo»«mo»§#215;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«msup»«mn»10«/mn»«mn»16«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«/mtable»«/math»
Please click here for a refresher on how to do unit conversions.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 2: Rearrange the formula.
Q = mcΔt
We need to isolate m. To do this, we need to move c and Δt to the other side of the formula. Since they are being multiplied, we use division to move them to the other side.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mi»Q«/mi»«mrow»«mi»c«/mi»«mo»§#8710;«/mo»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mi»m«/mi»«/math»
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses.Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«mrow»«mi»c«/mi»«mo»§#8710;«/mo»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»74«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»16«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«mstyle displaystyle=¨true¨»«mfenced»«mrow»«mn»1«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mstyle»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 4: Calculate the answer.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»74«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»16«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«mstyle displaystyle=¨true¨»«mfenced»«mrow»«mn»1«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mstyle»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»74«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»16«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»4«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo».«/mo»«mn»369«/mn»«mo»§#160;«/mo»«mn»9«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»16«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«/mtable»«/math»
The mass needs to be rounded to two significant digits, as two digits is the smallest number of digits found in the question.
m =1.4 x 1016 g
There is 1.4 x 1016 g of water in Lesser Slave Lake. - Cast iron is often used in cookware due to its non-stick properties and durability. A 1.5 kg cast iron skillet is heated from 21 ˚C to 85 ˚C using 44.2 kJ of thermal energy. What is the specific heat capacity of cast iron?
Step 1: List your variables.
Q = 44.2 kJ = 44 200 J
Note: The quantity of thermal energy must be converted to J to be used in the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mtable columnspacing=¨0px¨ columnalign=¨right center left¨» «mtr» «mtd» «mn»1«/mn» «mo»§#160;«/mo» «mn»000«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mn»1«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mtd» «/mtr» «mtr» «mtd» «mfrac» «mrow» «mn»1«/mn» «mo»§#160;«/mo» «mn»000«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mrow» «mrow» «mn»1«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «/mfrac» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mfrac» «mo»?«/mo» «mrow» «mn»44«/mn» «mo».«/mo» «mn»2«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «/mfrac» «/mtd» «/mtr» «/mtable» «/math»
Now cross-multiply and divide.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mtable columnspacing=¨0px¨ columnalign=¨right center left¨» «mtr» «mtd» «mn»1«/mn» «mo»§#160;«/mo» «mn»000«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «mo»§#215;«/mo» «mn»44«/mn» «mo».«/mo» «mn»2«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mn»44«/mn» «mo»§#160;«/mo» «mn»200«/mn» «/mtd» «/mtr» «mtr» «mtd» «mn»44«/mn» «mo»§#160;«/mo» «mn»200«/mn» «mo»/«/mo» «mn»1«/mn» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mn»44«/mn» «mo»§#160;«/mo» «mn»200«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mtd» «/mtr» «/mtable» «/math»
Please click here for a refresher on how to do unit conversions.
m = 1.5 kg = 1 500 g
Note: The mass must be converted to g to be used in the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mo»?«/mo»«mrow»«mn»1«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Now cross-multiply and divide.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#215;«/mo»«mn»1«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«/mtd»«/mtr»«mtr»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»/«/mo»«mn»1«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«mo»§#8710;«/mo»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»85«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«mo»-«/mo»«mn»21«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«mo»=«/mo»«mn»64«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 2: Rearrange the formula.
Q = mcΔt
We need to isolate c. To do this, we need to move m and Δt to the other side of the formula. Since they are being multiplied, we use division to move them to the other side.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mi»Q«/mi»«mrow»«mi»m«/mi»«mo»§#8710;«/mo»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mi»c«/mi»«/math»
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses.Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«mrow»«mi»m«/mi»«mo»§#8710;«/mo»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»c«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»44«/mn»«mo»§#160;«/mo»«mn»200«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mstyle displaystyle=¨true¨»«mfenced»«mrow»«mn»64«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mstyle»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»c«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 4: Calculate the answer.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»44«/mn»«mo»§#160;«/mo»«mn»200«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mfenced»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»64«/mn»«mo»§#160;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mrow»«/mfenced»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»c«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»44«/mn»«mo»§#160;«/mo»«mn»200«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»96«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»c«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»c«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»460«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#8226;«/mo»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
The specific heat capacity needs to be rounded to two significant digits, as two digits is the smallest number of digits found in the question.
c = 0.46 J/g•˚C
The specific heat capacity of cast iron is 0.46 J/g•˚C
Hydrologic Cycle
What is the hydrologic cycle, and how does it affect climate?
The hydrologic cycle is another term for the water cycle.
The water cycle follows the path of water as it evaporates into the atmosphere to become water vapour and clouds and then returns to the surface as precipitation to fill the lakes, rivers, and oceans or to seep into the soil to be used by plants. The water in the lakes, rivers, and oceans then evaporates back into the atmosphere to continue the cycle. The water that is absorbed by plants is then released back into the atmosphere through transpiration.
The water cycle also follows the path of the water that has been frozen in glaciers and is now melting.
D4.10 The hydrologic (water) cycle
The water cycle plays a large part in moving water through the biosphere. To do this, water molecules undergo many changes in phase (state). Water changes from its liquid state in rivers, lakes, and oceans into water vapour when it enters the atmosphere. It then changes back to its liquid state to form clouds and rain or its solid state to form snow. When glaciers and snow are melting, the water is changing from its solid state to either its liquid or gas state.
Did You Know?
D4.11 Fresh cob of ripe corn
A cornfield 2.5 acres in size can release as much as 37 000 L of water through transpiration each day. In comparison, an average adult human will exhale 0.35 L of water each day.
Every time water changes state, it is either releasing or absorbing thermal energy. When water is changing state, it is not changing temperature, instead the thermal energy is either being absorbed as bonds break (moving from a solid to a liquid or a gas) or is being released when bonds are formed (moving from a gas to a liquid or a solid). In this way, the hydrologic cycle is also transferring thermal energy throughout the biosphere. Since water is constantly changing phase in the hydrologic cycle and the water is not changing temperature while this is happening, it helps to absorb any excess thermal energy on Earth. This, in turn, helps to keep Earth’s temperature stable, since the thermal energy being used in the phase changes is not being used to raise the temperature of other parts of the biosphere.
The phase changes that occur in the hydrosphere play a huge role in the global transfer of thermal energy. When water evaporates in an area, it is absorbing 40 650 J of energy per mol. Often, it will then move to a different area and release thermal energy. In this way, it moves large amounts of thermal energy around Earth.
The phase changes that occur in the hydrosphere play a huge role in the global transfer of thermal energy. When water evaporates in an area, it is absorbing 40 650 J of energy per mol. Often, it will then move to a different area and release thermal energy. In this way, it moves large amounts of thermal energy around Earth.
D4.12 Fog can occur when a large amount of water has evaporated
Read This
Please read pages 382 to 383, and 389 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on what they hydrologic cycle is and the part it plays in transferring thermal energy around Earth. Remember, if you have any questions or you do not understand something, ask your teacher!
Practice Questions
Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.
-
Explain why 100.0 g of liquid water at 100 ˚C contains less thermal energy than 100.0 g of water vapour at 100 ˚C.
While both have the same temperature, the water vapour would contain more thermal energy than the liquid water. This is because thermal energy is absorbed during the breaking of bonds that happens when a liquid changes state to a gas. The water vapour has more thermal energy because those bonds have been broken and the thermal energy has been absorbed. The energy has been stored in the bonds of the liquid water.
- How does the hydrologic cycle move thermal energy around the biosphere?
The hydrologic cycle moves energy around the Earth because as the water evaporates into the atmosphere, it is breaking bonds and absorbing thermal energy. It then moves to a different area or part of the biosphere and changes back to a solid or liquid state in the form of precipitation and clouds. As water moves back into a solid or a liquid, bonds are being formed; this releases thermal energy into the surrounding area. In this way, thermal energy from one area or part of the biosphere was moved to a different area.
Heat of Fusion and Heat of Vaporization
How much energy is released during phase changes?
The heat of fusion of a substance tells us how much
thermal energy is absorbed when 1 mol of that substance changes from a solid state to a liquid state. The heat of solidification is the reverse phase change.
It tells us how much thermal energy is released when 1 mol of a substance changes from a liquid to a solid. Each phase change has its own term for how much energy is released or absorbed when 1 mol of a substance undergoes that phase change. It is important to note that each of these terms only applies as long as the temperature of the substance is not changing.
| Term | Phase Change
|
Energy Released or Absorbed
|
Amount of Energy (kJ/mol)
|
|---|---|---|---|
| heat of fusion
|
solid to liquid
|
absorbed | 6.01 |
| heat of solidification
|
liquid to a solid
|
released | 6.01 |
| heat of vaporization
|
liquid to a gas
|
absorbed | 40.66 |
| heat of condensation
|
gas to a liquid
|
released | 40.66 |
D4.13 Phase changes
The graph in image D4.14 shows the heating curve of water. It is important to note that every time the water changes phase, the graph becomes a horizontal line. This is because the temperature of water is not changing while it changes phases; instead, the thermal energy is either being released or absorbed as needed to break or form bonds.
D4.14 Heating curve of water
Read This
Please read pages 383 to 385 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the new terms presented and the heating curve of water presented. Remember, if you have any questions or you do not understand something, ask your teacher!
Practice Questions
Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.
- What is the difference between the heat of vaporization and the heat of condensation?
The heat of vaporization is a measure of the amount of thermal energy absorbed by 1 mol of a substance when it changes from a liquid to a gas; while the heat of condensation is the amount of thermal energy released while the substance changes from a gas to a liquid.
- What happens to the temperature of water as it changes phases? Why does this happen?
The temperature does not change as water changes phases. This is because the thermal energy is being absorbed or released as part of the phase change, rather than being used to increase the temperature of the water. Another possible answer is the thermal energy is not being used to increase the kinetic energy of the particles during a phase change. Instead, the thermal energy is being absorbed or released by the forces between the particles, causing them to break or form.
Virtual Lab
Phase Changes © Explore Learning
Background Information:
This lab will help you visualize what happens to the particles of water during the different phases, as well as what happens during a phase change. It will also help create your own heating curve for water.
Please note: if you scroll down while in the Gizmo you will see a list of questions. You DO NOT need to complete these questions. You are able to complete them for extra practice if you would like.
Parts of this lab will be used in 4.3 Assignment
This lab will help you visualize what happens to the particles of water during the different phases, as well as what happens during a phase change. It will also help create your own heating curve for water.
Please note: if you scroll down while in the Gizmo you will see a list of questions. You DO NOT need to complete these questions. You are able to complete them for extra practice if you would like.
Parts of this lab will be used in 4.3 Assignment
- Click on the play icon to open the Gizmo. Print students can access the Gizmo from the Online Resources for Print Students section in their online course.
- Change the initial water temperature to –10 ˚C by using the slider on the right side of the screen or by typing in a new temperature.
- Click on “Micro view” at the bottom right side of the screen.
- Slide the heat energy slider all the way to the right.
- Click on the play button.
© Explore learning
D4.15 Initial settings
D4.15 Initial settings
- Observe what happens to the molecules of water as they change phases and during each phase.
- Note what happens to the temperature as the water changes from a solid to a liquid and a liquid to a gas.
- Once you have watched the water change from a solid to a liquid to a gas, press pause on the simulation and switch to the “GRAPH” tab.
- Observe the heating curve of water. You may need to adjust the zoom to see your graph better. To do this, click the negative button on the graph. Take a screen shot or sketch the heating curve.. Save the screen shot as it will be submitted for a mark on 4.3 Assignment
- Click reset and try running the simulation backward. Set the initial temperature of the water at 120 ˚C and slide the heat energy slider all the way to the left. Click play and watch what happens to the cooling curve of water. Take a screen shot or sketch the cooling curve for your notes.
- Please return to the top of this page and click on analysis to complete the analysis questions.
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|
|
Initial Temperature (°C)
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Final Temperature (°C)
|
Time It Took for Temperatures
to Reach Equilibrium |
|---|---|---|---|---|
| Aluminum | beaker A
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95
|
50
|
22 min 50 s
|
|
beaker B
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5
|
50
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||
| Copper | beaker A |
|
|
|
| beaker B
|
|
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||
| Steel | beaker A |
|
|
|
| beaker B
|
|
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||
| Glass | beaker A |
|
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| beaker B
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- What is different about the molecules of water during the solid, liquid, and gas phases?
As the molecules move from a solid to a liquid to a gas, they get more motion and the bonds between them break. As a solid, the molecules are close together and do not move around much; they just vibrate. As a liquid, the molecules move around more. As a gas, the molecules have lots of space between them and they move around quickly.
- What happens to the temperature of the water as it changes phases? Why do you observe this?
As the phase changes, the temperature stays stable. This is because the thermal energy is being used to break or create bonds, rather than being used to raise the temperature of the water.
- What happens to the temperature of the water once the phase change is completed? Why do you observe this?
The temperature of the water will begin to increase again once the phase change is completed. This is because the thermal energy is no longer being used to change the bonds and so can now go toward increasing the temperature.
- Which of the following descriptions explains the process of water freezing?
- Water molecules speed up until they escape into a different phase.
- Water molecules slow down until they begin to become fixed in position.
- Water molecules are squeezed together by high pressure until they begin to stick together.
- As water molecules cool, their shape changes from rounded to irregular, making them more likely to link to one another.
Calculating Heat of Fusion and Heat of Vaporization
The amount of thermal energy being released or absorbed by a phase change can be calculated using a simple formula.
Using the quantity of thermal energy that was added and the number of moles of the substance, you can calculate the heat of fusion or the heat of vaporization of a substance.
D4.16 Ice undergoing fusion
«math»
«msub mathcolor=¨#FFFFFF¨»
«mi mathcolor=¨#FFFFFF¨»H«/mi»
«mrow»
«mi»f«/mi»
«mi»u«/mi»
«mi»s«/mi»
«/mrow»
«/msub»
«mo mathcolor=¨#FFFFFF¨»=«/mo»
«mfrac mathcolor=¨#FFFFFF¨»
«mi»Q«/mi»
«mi»n«/mi»
«/mfrac»
«/math»
Hfus = heat of fusion in kJ/mol (This can also be Hvap depending on if you are calculating the heat of fusion or heat of vaporization.)
Q = quantity of thermal energy added in kJ (This is the same Q as seen in the specific heat capacity formula, however, it uses a different unit.)
n = the amount of the substance in mol
Hfus = heat of fusion in kJ/mol (This can also be Hvap depending on if you are calculating the heat of fusion or heat of vaporization.)
Q = quantity of thermal energy added in kJ (This is the same Q as seen in the specific heat capacity formula, however, it uses a different unit.)
n = the amount of the substance in mol
You can use this formula to determine
- the heat of fusion or the heat of vaporization
- the quantity of thermal energy added
- the amount of the substance used
Let’s try some examples. Each example has a video to go with it. To play the video, click on the play icon next to the example.
Digging Deeper
D4.17 Ice floating in a drink
Normally, a solid object is denser than a liquid one. That is not the case with water: Ice is actually less dense than water, which is why ice floats in our drinks instead of sinking. Go to the following link to learn more about the density of ice.
Learn More
Examples
When 45.0 kJ of thermal energy is added to 7.49 mol of water, the ice melts. What is the heat of fusion of water?
https://adlc.wistia.com/medias/m7elgyguzs
Step 1: List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»45«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»7«/mn»«mo».«/mo»«mn»49«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»45«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»7«/mn»«mo».«/mo»«mn»49«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Substitute the values into the formula.
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mrow» «mn»45«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «mrow» «mn»7«/mn» «mo».«/mo» «mn»49«/mn» «mo»§#160;«/mo» «mi»mol«/mi» «/mrow» «/mfrac» «/math»
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mrow» «mn»45«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «mrow» «mn»7«/mn» «mo».«/mo» «mn»49«/mn» «mo»§#160;«/mo» «mi»mol«/mi» «/mrow» «/mfrac» «/math»
Step 3: Calculate the answer.
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mrow» «mn»45«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «mrow» «mn»7«/mn» «mo».«/mo» «mn»49«/mn» «mo»§#160;«/mo» «mi»mol«/mi» «/mrow» «/mfrac» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mn»6«/mn» «mo».«/mo» «mn»008«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «mo»/«/mo» «mi»mol«/mi» «/math»
The heat of fusion needs to be rounded to three significant digits, as three digits is the smallest number of digits found in the question.
Hfus = 6.01 kJ/mol
The heat of fusion of water is 6.01 kJ/mol.
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mrow» «mn»45«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «mrow» «mn»7«/mn» «mo».«/mo» «mn»49«/mn» «mo»§#160;«/mo» «mi»mol«/mi» «/mrow» «/mfrac» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mn»6«/mn» «mo».«/mo» «mn»008«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «mo»/«/mo» «mi»mol«/mi» «/math»
The heat of fusion needs to be rounded to three significant digits, as three digits is the smallest number of digits found in the question.
Hfus = 6.01 kJ/mol
The heat of fusion of water is 6.01 kJ/mol.
A pot containing 100.0 g of water is left outside on a hot day. The sun provides 225.65 kJ of thermal energy, which causes the water to evaporate. What is the heat of vaporization of the water?
https://adlc.wistia.com/medias/62lxxu5v41
Step 1: List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»225«/mn»«mo».«/mo»«mn»65«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»100«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«/mtable»«/math»
Note: This question gives you the amount of water in g rather than mol. You will need to convert the mass to mol using the following formula:
«math» «mi»n«/mi» «mo»=«/mo» «mfrac» «mi»m«/mi» «mi»M«/mi» «/mfrac» «/math»
You may recognize this formula from your chemistry unit. For a review on how to use it, click here.
Remember from your chemistry unit that the molar mass or M of water is 18.02 g/mol. For a review on how to calculate molar mass, click here.
«math»«mi»n«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»100«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»18«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«mi»n«/mi»«mo»=«/mo»«mn»5«/mn»«mo».«/mo»«mn»549«/mn»«mo»§#160;«/mo»«mn»389«/mn»«mo»§#160;«/mo»«mn»567«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/math»
Do not round this number when calculating heat of vaporization.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»225«/mn»«mo».«/mo»«mn»65«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»100«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«/mtable»«/math»
Note: This question gives you the amount of water in g rather than mol. You will need to convert the mass to mol using the following formula:
«math» «mi»n«/mi» «mo»=«/mo» «mfrac» «mi»m«/mi» «mi»M«/mi» «/mfrac» «/math»
You may recognize this formula from your chemistry unit. For a review on how to use it, click here.
Remember from your chemistry unit that the molar mass or M of water is 18.02 g/mol. For a review on how to calculate molar mass, click here.
«math»«mi»n«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»100«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»18«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«mi»n«/mi»«mo»=«/mo»«mn»5«/mn»«mo».«/mo»«mn»549«/mn»«mo»§#160;«/mo»«mn»389«/mn»«mo»§#160;«/mo»«mn»567«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/math»
Do not round this number when calculating heat of vaporization.
Step 2: Substitute the values into the formula.
«math» «msub» «mi»H«/mi» «mrow» «mi»v«/mi» «mi»a«/mi» «mi»p«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»H«/mi» «mrow» «mi»v«/mi» «mi»a«/mi» «mi»p«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mrow» «mn»225«/mn» «mo».«/mo» «mn»65«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «mrow» «mn»5«/mn» «mo».«/mo» «mn»549«/mn» «mo»§#160;«/mo» «mn»389«/mn» «mo»§#160;«/mo» «mn»567«/mn» «mo»§#160;«/mo» «mi»mol«/mi» «/mrow» «/mfrac» «/math»
«math» «msub» «mi»H«/mi» «mrow» «mi»v«/mi» «mi»a«/mi» «mi»p«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»H«/mi» «mrow» «mi»v«/mi» «mi»a«/mi» «mi»p«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mrow» «mn»225«/mn» «mo».«/mo» «mn»65«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «mrow» «mn»5«/mn» «mo».«/mo» «mn»549«/mn» «mo»§#160;«/mo» «mn»389«/mn» «mo»§#160;«/mo» «mn»567«/mn» «mo»§#160;«/mo» «mi»mol«/mi» «/mrow» «/mfrac» «/math»
Step 3: Calculate the answer.
«math»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mrow»«mn»225«/mn»«mo».«/mo»«mn»65«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»5«/mn»«mo».«/mo»«mn»549«/mn»«mo»§#160;«/mo»«mn»389«/mn»«mo»§#160;«/mo»«mn»567«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»40«/mn»«mo».«/mo»«mn»662«/mn»«mo»§#160;«/mo»«mn»13«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
The heat of vaporization needs to be rounded to four significant digits, as four digits is the smallest number of digits found in the question.
Hvap=40.66 kJ/mol
The heat of vaporization of water is 40.66 kJ/mol.
«math»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mrow»«mn»225«/mn»«mo».«/mo»«mn»65«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»5«/mn»«mo».«/mo»«mn»549«/mn»«mo»§#160;«/mo»«mn»389«/mn»«mo»§#160;«/mo»«mn»567«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»40«/mn»«mo».«/mo»«mn»662«/mn»«mo»§#160;«/mo»«mn»13«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
The heat of vaporization needs to be rounded to four significant digits, as four digits is the smallest number of digits found in the question.
Hvap=40.66 kJ/mol
The heat of vaporization of water is 40.66 kJ/mol.
D4.18 Nugget of gold
A jewellery maker wants to melt some gold to make a ring. If he has 0.38 mol of gold and he knows gold has a heat of fusion of 13.2 kJ/mol, how much thermal energy must he add?
https://adlc.wistia.com/medias/t1yrk8sh9r
Step 1: List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»13«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»38«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»13«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»38«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Rearrange the formula.
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «/math»
We need to isolate Q. To do this, we need to move n to the other side of the formula. Since it is being divided, we use multiplication to move it to the other side.
Q = Hfusn
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses!
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «/math»
We need to isolate Q. To do this, we need to move n to the other side of the formula. Since it is being divided, we use multiplication to move it to the other side.
Q = Hfusn
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses!
Step 3: Substitute the values into the formula.
Q = Hfusn
Q = (13.2 kJ/mol)(0.38 mol)
Q = Hfusn
Q = (13.2 kJ/mol)(0.38 mol)
Step 4: Calculate the answer.
Q = (13.2 kJ/mol)(0.38 mol)
Q = 5.016 kJ
This answer needs to be rounded to two significant digits, as two digits is the smallest number of digits in the question. Please note that 0.38 has two digits as the zero in front does not count.
Q = 5.0 kJ
The jewellery maker would need to add 5.0 kJ of energy to melt 0.38 mol of gold.
Q = (13.2 kJ/mol)(0.38 mol)
Q = 5.016 kJ
This answer needs to be rounded to two significant digits, as two digits is the smallest number of digits in the question. Please note that 0.38 has two digits as the zero in front does not count.
Q = 5.0 kJ
The jewellery maker would need to add 5.0 kJ of energy to melt 0.38 mol of gold.
The same jewellery maker decides to make a ring out of silver instead. He already has the 5.0 kJ of thermal energy to add, and he knows silver has a heat of fusion of 11.3 kJ/mol. How many mol of silver can he melt?
https://adlc.wistia.com/medias/pk45ls6cfz
Step 1: List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»11«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»11«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Rearrange the formula.
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «/math»
We need to isolate n. To do this, we need to move n to the other side of the formula, since we cannot isolate a variable on the bottom of a fraction. This means we need to get rid of the fraction. Since it is being divided, we use multiplication to move it to the other side.
Q = Hfusn
Now that we have gotten rid of the fraction, we can move the Hfus to the other side to isolate n. To do this we, need to divide, since Hfus is being multiplied.
«math» «mfrac» «mi»Q«/mi» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «/mfrac» «mo»=«/mo» «mi»n«/mi» «/math»
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses!
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «/math»
We need to isolate n. To do this, we need to move n to the other side of the formula, since we cannot isolate a variable on the bottom of a fraction. This means we need to get rid of the fraction. Since it is being divided, we use multiplication to move it to the other side.
Q = Hfusn
Now that we have gotten rid of the fraction, we can move the Hfus to the other side to isolate n. To do this we, need to divide, since Hfus is being multiplied.
«math» «mfrac» «mi»Q«/mi» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «/mfrac» «mo»=«/mo» «mi»n«/mi» «/math»
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses!
Step 3: Substitute the values into the formula.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»11«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»11«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»11«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»442«/mn»«mo»§#160;«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
This answer needs to be rounded to two significant digits, as two digits is the smallest number of digits found in the question.
n = 0.44 mol
The jewellery maker can melt 0.44 mol of silver with 5.0 kJ of thermal energy.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»11«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»442«/mn»«mo»§#160;«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
This answer needs to be rounded to two significant digits, as two digits is the smallest number of digits found in the question.
n = 0.44 mol
The jewellery maker can melt 0.44 mol of silver with 5.0 kJ of thermal energy.
Read This
Please read pages 385 to 387 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the different calculations you can do with the heat of vaporization and the heat of fusion formulas and how to do them. Please also try the practice questions found on these pages. Remember, if you have any questions or you do not understand something, ask your teacher!
Practice Questions
Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.
- Spring time in Alberta means lots of melting snow and ice, but how much thermal energy does this absorb from the surrounding environment? If 1 500 g of snow (known to have a heat of fusion of 6.01 kJ/mol) melts, how much thermal energy is absorbed?
Step 1: List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»f«/mi»«mi»u«/mi»«mi»s«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«/mtable»«/math»
Note: This question gives you the amount of water in g rather than mol. You will need to convert the mass to mol using the following formula:
«math» «mi»n«/mi» «mo»=«/mo» «mfrac» «mi»m«/mi» «mi»M«/mi» «/mfrac» «/math»
You may recognize this formula from your chemistry unit. For a review on how to use it, click here.
Remember from your chemistry unit that the molar mass or M of water is 18.02 g/mol. For a review on how to calculate molar mass, click here.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»18«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»83«/mn»«mo».«/mo»«mn»240«/mn»«mo»§#160;«/mo»«mn»843«/mn»«mo»§#160;«/mo»«mn»51«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Do not round this number when calculating the quantity of thermal energy.
Step 2: Rearrange the formula.
«math» «msub» «mi»H«/mi» «mrow» «mi»f«/mi» «mi»u«/mi» «mi»s«/mi» «/mrow» «/msub» «mo»=«/mo» «mfrac» «mi»Q«/mi» «mi»n«/mi» «/mfrac» «/math»
We need to isolate Q. To do this, we need to move n to the other side of the formula. Since it is being divided, we use multiplication to move it to the other side.
Q = Hfusn
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses!
Step 3: Substitute the values into the formula.
Q = Hfusn
Q = (6.01 kJ/mol)(83.240 843 51 mol)
Step 4: Calculate the answer.
Q = (6.01 kJ/mol)(83.240 843 51 mol)
Q = 500.277 kJ
The heat of vaporization needs to be rounded to three significant digits, as three digits is the smallest number of digits found in the question.
Q = 500 kJ
The amount of thermal energy absorbed is 500 kJ.
- Sylvan lake near Red Deer, Alberta, is a popular summer hang out spot for people of all ages. It can get quite warm in the summer, which causes the water in the lake to evaporate into the atmosphere. During dry years, the level of the lake can drop. If the sun provides 1 500 kJ of thermal energy to be absorbed by the water molecules and the heat of vaporization of water is 40.66 kJ/mol.
- How many mol of water is evaporated?Step 1: List the variables
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»40«/mn»«mo».«/mo»«mn»66«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»Q«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»Step 2: Rearrange the formula.
«math»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mi»Q«/mi»«mi»n«/mi»«/mfrac»«/math»
We need to isolate n. To do this, we need to move n to the other side of the formula, since we cannot isolate a variable on the bottom of a fraction. This means we need to get rid of the fraction. Since it is being divided, we use multiplication to move it to the other side.
Q = Hvapn
Now that we have gotten rid of the fraction, we can move the Hvap to the other side to isolate n. To do this, we need to divide, since Hvap is being multiplied.
«math»«mfrac»«mi»Q«/mi»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«/mfrac»«mo»=«/mo»«mi»n«/mi»«/math»
If you struggle with rearranging formulas, please contact your teacher for help. This is an important skill that will be used in all of your future math and science courses!Step 3: Substitute the values into the formula.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»Q«/mi»«msub»«mi»H«/mi»«mrow»«mi»v«/mi»«mi»a«/mi»«mi»p«/mi»«/mrow»«/msub»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»40«/mn»«mo».«/mo»«mn»66«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 4: Calculate the answer.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mo»§#160;«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mn»500«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»40«/mn»«mo».«/mo»«mn»66«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»n«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»36«/mn»«mo».«/mo»«mn»891«/mn»«mo»§#160;«/mo»«mn»293«/mn»«mo»§#160;«/mo»«mn»65«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
n = 36.891 293 65 mol
This answer needs to be rounded to four significant digits, as four digits is the smallest number of digits found in the question.
n = 36.89 mol
There will be 36.89 mol of water evaporated from Sylvan Lake. - How many g of water is evaporated?To convert mol to g, you need to use the following formula from the chemistry unit:
«math»«mi»n«/mi»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/math»Step 1: List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»n«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»36«/mn»«mo».«/mo»«mn»89«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»M«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»18«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 2: Rearrange the formula.
«math»«mi»n«/mi»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/math»
We need to isolate m. To do this, we need to move M to the other side of the formula. Since it is being divided, we use multiplication to move it to the other side.
nM = mStep 3: Substitute the values into the formula.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»n«/mi»«mi»M«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«mrow»«mo»(«/mo»«mn»36«/mn»«mo».«/mo»«mn»89«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»(«/mo»«mn»18«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«/mrow»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»664«/mn»«mo».«/mo»«mn»757«/mn»«mo»§#160;«/mo»«mn»8«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«/mtable»«/math»
This answer needs to be rounded to four significant digits, as four digits is the smallest number of digits found in the question.
m = 664.8 g
664.8 g of water would evaporate from Sylvan Lake.
- How many mol of water is evaporated?
Thermal Energy and the Hydrologic Cycle
The presence of large bodies of water plays a huge role on the climate in the areas surrounding them.
D4.19 Clouds travelling over North America
Due to its high specific heat capacity, water is able to absorb, store, and release large amounts of thermal energy without affecting its own temperature much. This means that areas around large bodies of water have much less variation in their climate. The water is able to release thermal energy when the temperature starts to get very cold and absorb thermal energy when the temperature gets very warm. It can also create warmer or colder climates depending on if it is a warmer or colder body of water.
Due to its strong hydrogen bonds, water has relatively high heats of fusion and vaporization. This means that water absorbs a large amount of thermal energy when changing from a solid to a liquid or a liquid to a gas. This contributes to the movement of thermal energy around Earth. As the water evaporates into the atmosphere, it absorbs a large amount of thermal energy. This water vapour then travels to a different area of Earth where the water changes back to a liquid or a solid to form precipitation. This releases a large amount of thermal energy into that area of Earth.
In the next lesson, we will take what was discussed in this lesson and apply it to global wind patterns and ocean currents to see how they affect climate.
Due to its strong hydrogen bonds, water has relatively high heats of fusion and vaporization. This means that water absorbs a large amount of thermal energy when changing from a solid to a liquid or a liquid to a gas. This contributes to the movement of thermal energy around Earth. As the water evaporates into the atmosphere, it absorbs a large amount of thermal energy. This water vapour then travels to a different area of Earth where the water changes back to a liquid or a solid to form precipitation. This releases a large amount of thermal energy into that area of Earth.
In the next lesson, we will take what was discussed in this lesson and apply it to global wind patterns and ocean currents to see how they affect climate.
4.3 Assignment
Unit 4 Formative Assessment Lesson 2
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4.3 Assignment