Module 4

Lesson 1.2  Using Calorimetry to Find Q



Key Concepts


Calorimetry is the technological process of measuring energy changes in an isolated system. An isolated system is a system in which neither matter nor energy can move in or out.

A calorimeter is an insulated container that contains a measured quantity of water and a thermometer. Calorimeters can be high-tech (bomb calorimeters) or they can be very simple, consisting of a polystyrene container with a lid, a measured quantity of water, and a thermometer.

Calorimetry is based on two scientific principles, namely the first and second laws of thermodynamics. These laws state the following:

  • Energy can be neither created nor destroyed.

  • Heat transfers naturally from warmer to cooler objects.

According to these laws, energy lost by one substance must equal the energy gained by another substance. This concept is at the heart of calorimetry.

When a chemical reaction takes place in a calorimeter, thermal energy (or "heat") is either absorbed from the surroundings or released to the surroundings. This energy, symbolized as "Q", causes a temperature change in the water surrounding the chemical system. If the reaction occurring in the calorimeter is exothermic, the temperature of the surrounding water increases. Conversely, if the reaction occurring in the calorimeter is endothermic, the temperature of the surrounding water decreases.

Learning Tip

Thermal energy (Q) represents the total kinetic energy of the entities of a substance (such as molecules of water in a calorimeter).

The following equation is used to find Q, the total amount of thermal energy transferred in a reaction:

Q = mcΔt

m is the mass (g) of the water or solution in the calorimeter.

c is the specific heat capacity of the water or solution in the calorimeter (J/g°C).

Δt is the temperature change in the water or solution surrounding the chemical system (°C).

The Specific Heat Capacity of a substance is the quantity of energy required to raise the temperature of a unit mass of a substance by one degree C or one degree Kelvin (J/g°C).



When working out calorimetry problems, assume the following:

  • Unless otherwise stated, the amount of thermal energy gained by all components of the calorimeter (container, lid, thermometer, stirrer) is negligible and, therefore, can be ignored.

  • Calorimeters are not 100% efficient. That is, there is always some heat lost to the calorimeter and/or surroundings. However, for polystyrene containers, consider this heat loss to be negligible.

  • Dilute aqueous solutions have the same density as water (1 g/mL) and the same specific heat capacity as water (4.19 J/g°C).


Watch


The Fuse School - Global Education

 

Virtual Investigation

 Study the photographs below to better understand how calorimetry can be used to find Q. Answer the associated questions in your Module 4 Summative Assessment.



Step 1: The mass of the nut (pecan) is measured.





Step 2: A paper clip is bent to form a stand that will support a nut above the lab bench.


Step 3:  50mL of tap water is placed in the tin can.




Step 4:  The temperature of the water in the can is measured. 

 

Step 5: A can of water is suspended inside the ring and above the nut by putting  a pencil through the holes and under the rim of the can.

Step 6: The nut is lit .
Step 7: The can containing the water is positioned above the burning nut.

Step 8: When the nut has stopped burning, the final temperature of the water in the can is measured.





Check Your Understanding


Complete "Practice" questions 4 to 8 on page 487 of the textbook. Check your answers by clicking the banner below.

 
Page 487 Practice Question 4
  1. If the mass of material undergoing temperature change is doubled and the other variables remain the same, then the thermal energy is doubled.
  2. If the specific heat capacity of material undergoing temperature change is divided by two (halved) and the other variables remain the same, then the thermal energy is halved.
  3. If the temperature change of the material is tripled and the other variables remain the same, then the thermal energy is tripled.
  4. If all three variables (m, c, and Δt) are doubled, then the thermal energy is increased by a factor of 8 (2 x 2 x 2 = 8).

Page 487 Practice Question 5

The values in "Table 1" on page 486 of the textbook show that iron has a lower specific heat capacity than aluminium has. Therefore, iron requires less thermal energy to change temperature by one degree. As a result of its lower specific heat capacity, a 100-g iron mass would undergo a greater temperature change than a 100-g aluminium mass if the same quantity of energy was applied to both samples.


Page 487 Practice Question 6

\( \mathrm { Q = mc\Delta t } \)

\( \mathrm { = 250 mL \times \dfrac{1~g}{1~mL} \times 4.19 \frac{J}{g°C} \times (95°C - 15°C) = 83800~J } \)

\( \mathrm { = 84~kJ } \)


Page 487 Practice Question 7

\( \mathrm { Q = mc\Delta t } \)

\( \mathrm { = 1.0 L \times \dfrac{1~kg}{1~L} \times \dfrac{4.19~kJ}{kg°C} \times (97°C - 5.0°C) } \)

\( \mathrm { = 385.48~kJ } \)

\( \mathrm { = 3.9 \times 10^2~kJ } \)


Page 487 Practice Question 8

\( \mathrm { Q = mc\Delta t } \)

\( \mathrm { = 150~mL \times \dfrac{1~g}{1~mL} \times \dfrac{1~kg}{1000~g} \times \dfrac{4.19~kJ}{kg°C} \times (52.8°C- 20.6°C) } \)

\( \mathrm { = 20.238 kJ } \)

\( \mathrm { \%~Efficiency = \dfrac{20.238~kJ}{30.0~kJ} \times 100 = 67.5\% } \)

\( \mathrm { The~calorimeter~is~67.5\%~efficient. } \)