10.5 – Heat Capacity & Specific Heat Capacity

Different bodies have different capacities to store internal energy. Hence we need to know about the heat capacity and/or specific heat capacity of the material.

Heat Capacity, C

Example 6
100 g of water requires 12 600 J of thermal energy to raise it from 30 °C to 60 °C.
(a) Calculate the heat capacity of 100 g of water.

 

C = Q/Δθ

C = 12 600 / (60 – 30)

C = 420 J °C-1

 

(b) Calculate the heat capacity of 1000 g of water.

10× the mass of water will require 10× the energy to raise the temperature by the same amount.

C = 10× 420

C = 4200 J °C-1

 

Specific Heat Capacity, c

  • Heat capacity C may vary from body to body (e.g. of same substance but different mass). It is difficult to compare which materials have higher or lower heat capacities unless we compare standard masses of these materials.
  • Hence, we use heat capacity per unit mass or specific heat capacity c.

c = C / m.

 

  • Specific heat capacity c is more commonly used because it is the fundamental property of a substance or material. It is different for different substances.

e.g. for water, c = 4200 J kg-1 K-1 or 4.2 J g-1 K-1

Caution: Units
Always check that the units of mass in m and in c are consistent in the equation:

E.g. Q = m c Δθ,       for copper:   use m = 0.150 kg and c = 400 J kg-1 K-1        

or        use m = 150 g and c = 0.40 J g-1 K-1

 

Example 7
12 000 J of thermal energy was supplied to a 1.00 kg of aluminium block. The temperature of the block then rises from 31.0 °C  to 45.0 °C.

Calculate the specific heat capacity of the aluminium block.

Q = m c Δθ

c    =  Q / (m Δθ) = 12 000 J / ((1.00 kg)(45 – 31)K)

c = 12000 / 14 = 857.14

c ≈ 857 J kg–1 K–1  (3 s.f.)   or 857 J kg-1 °C–1
Example 11
A heating coil with 230 V applied across and a current 8.6 A through it, is used to heat 2.0 kg of a liquid. It takes 5.0 minutes to raise the temperature of this liquid from 30 °C to 100 °C.

Assuming no energy is lost to the surroundings, determine the specific heat capacity of this liquid.

energy supplied by heating coil = energy gained by the liquid

Pt = m c Δθ

V I t = m c Δθ                        (Electrical power is given by E = V I t )

(230)(8.6)(5.0 × 60) = (2.0) c (100 – 30)

c = 593400 / 140

= 4239

≈ 4200 J (kg °C)-1 (2 sig.fig.)

Specific heat capacity of a single substance is different when in solid, liquid or gaseous state. Therefore, when considering specific heat capacity of a substance, it is necessary to take note of the physical state of the substance at that instant.

 

Equilibrium Temperature

  • The transfer of thermal energy into (or out of) a body can cause a change in temperature of the body.
  • When two objects at two different initial temperatures are brought together, thermal energy transfer will take place between them.
    • Thermal energy is transferred from a region of higher temperature to a region of lower temperature.
  • The thermal energy in the system can be said to be conserved.
    • Thermal energy lost by the initially hotter object is equal to the sum of the thermal energy gained by the colder object and the heat loss to the environment.
  • When two objects have reached the same temperature we say they are in thermal equilibrium.

 

Example 8

A brass cube of mass 0.20 kg at 100 °C is transferred into a container with negligible heat capacity containing 0.50 kg of paraffin at 30 °C. Calculate the final temperature of paraffin, assuming there is no energy loss to the surroundings.

Given:
     specific heat capacity of brass = 380 J kg–1 °C–1;
     specific heat capacity of paraffin = 2000 J kg–1 °C–1

When the two items are brought together heat flows from the hotter object into the colder object until they reach thermal equalibrium.

We can think of the heat transfers as two separate things:

Q1 being the heat lost by the brass cube as it cools, and Q2 being the heat gained by the paraffin as it warms.

Of course Q1 is equal to Q2 as the energy flows from the brass into the paraffin.

Energy lost by brass = energy gained by paraffin

mb cb (100 – θ) = mp cp (θ – 30)

0.20 x 380 x (100 – θ) = 0.50 x 2000 (θ – 30)

θ = 35 °C

 

Experimental Determination of Specific Heat Capacities (ENRICHMENT)

Experimental Determination of Specific Heat Capacity, c, of a Solid

example: Aluminium

Procedure

  1. Measure mass of substance, m
  2. Record reading on thermometer, θ1
  3. Switch on the heater
  4. Start the stopwatch
  5. After a certain time interval, record the reading on the thermometer again, θ2
  6. Switch off the heater
  7. Record the time interval (t) and power output (P) of heat supply OR record reading on joulemeter (Q) attached to heat supply

Measurements & Results

  • Mass of substance, m
  • Initial temperature of substance, θ1
  • Final temperature of substance, θ2
  • Total thermal energy supplied to the substance, Q

Q = P t = m c Δθ = m c (θ2 – θ1)

Precautions

  • The holes should not be too large so that there is better contact between the heater and thermometer with the aluminium block.

Sources of Error

  • There is loss of thermal energy to the surroundings.

Possible Improvements to the Set-up

  • The aluminium block can be lagged (insulated or covered with a thermal insulator)
Experimental Determination of Specific Heat Capacity, c, of a Liquid

example: Ethanol

Procedure is as above.

Precautions

  • Stir the ethanol to allow better distribution of heat within the ethanol.

Sources of Error

  • There is loss of thermal energy to the surroundings.

Possible Improvements to the Set-up

  • Reduce heat loss by lagging the container
Example
The following data was collected during an experiment to determine the specific heat capacity of ethanol.

Mass of ethanol: 0.25 kg

Joulemeter reading of heat supply: 5000 J
Initial temperature of ethanol: 35.0 °C
Final temperature of ethanol: 47.0 °C

Calculate the specific heat capacity of ethanol.

Energy supplied by heater = Energy gained by ethanol

E = m c Δθ

5000 J = (0.25 kg) c (47 – 35)

c =1667

c ≈ 1700J kg–1 K–1   (2 s.f.)
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