Lesson 4 Thermal Energy
Completion requirements
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