Heat Capacity and Specific Heat Capacity – SS2 Physics Lesson Note (Second Term)

Heat capacity is the quantity of heat needed to raise the temperature of a substance by one degree, measured in J/K. Specific heat capacity is the heat needed to raise the temperature of a unit mass of a substance by one degree, measured in J/kg K. The heat gained or lost by a body is given by Q = mcΔθ.

Behavioural objectives

By the end of the lesson, students should be able to:

1. define heat capacity and specific heat capacity and state their units;
2. use the equation Q = mcΔθ to find the quantity of heat gained or lost by a body;
3. describe how to determine specific heat capacity by the method of mixtures and by the electrical method;
4. solve simple problems on specific heat capacity.

How is heat energy measured?

To assess the quantity of heat energy possessed by a body, three quantities are needed:

1. the change in temperature (θ)
2. the specific heat capacity of the body (c)
3. the mass of the body (m)

The quantity of heat Q of a body is the product of these three quantities, as expressed by the equation:

Q = mcΔθ ……(1)

It is measured in joules (J).

What is heat capacity?

Heat capacity is the quantity of heat required to raise the temperature of a substance by one degree. It is measured in joules per kelvin (J/K).

C = mc ……(2)

What is specific heat capacity?

The specific heat capacity of a substance is the heat required to raise the temperature of a unit mass of the substance through a one-degree change in temperature.

The quantity of heat Q received by a body is proportional to its mass (m), to its temperature change, and to the nature of the material the body is made of. Thus:

Q = mcΔθ ……(3)

Q = mc(θ₂ – θ₁) ……(4)

Here c is a constant of proportionality called the specific heat capacity of the body. It depends on the nature of the body.

c = Q ÷ m(θ₂ – θ₁) ……(5)

The unit of specific heat capacity is J/kg K (J kg^-1 K^-1).

It can be determined by using:

1. the method of mixtures
2. the electrical method

How is specific heat capacity determined by the method of mixtures?

1. Weigh the solid lead block on a balance. Call its mass M_s.
2. Dry a lagged calorimeter and weigh it. Call its mass M_c.
3. Half fill the calorimeter with water and weigh it again. Call this mass M_cw. Take the initial temperature of the water as θ₁.
4. Suspend the lead block in boiling water at temperature θ₂.
5. Transfer the lead block to the calorimeter and stir the mixture to keep the temperature uniform. Call the final temperature θ₃.

The specific heat capacity of the lead can be calculated from the fact that heat lost by the lead = heat gained by the calorimeter and water.

Let the specific heat capacities of the lead, the calorimeter and the water be C_s, C_c and C_w respectively. The mass of water is M_cw – M_c.

Heat lost by the lead = heat gained by the calorimeter and water

M_sC_s(θ₂ – θ₃) = M_cC_c(θ₃ – θ₁) + (M_cw – M_c)C_w(θ₃ – θ₁) ……(6)

C_s = [M_cC_c(θ₃ – θ₁) + (M_cw – M_c)C_w(θ₃ – θ₁)] ÷ [M_s(θ₂ – θ₃)] ……(7)

How is specific heat capacity determined by the electrical method?

For a solid

To calculate the specific heat capacity C_b of a solid brass block, make two holes in a weighed brass block. Insert a thermometer in one hole and a heating element, connected to a power supply, in the other. Pour oil into the holes to ensure good thermal contact.

Assuming no heat is lost to the surroundings, the total electrical heat energy supplied by the coil is equal to the heat gained by the brass.

Heat energy supplied by the coil = heat gained by the brass

IVt = MC_bΔθ ……(8)

where I = current, V = voltage, t = time and M = mass of the brass block.

From Ohm’s law:

V = IR ……(9)

Substituting in equation 8:

I²Rt = MC_bΔθ ……(10)

V²t ÷ R = MC_bΔθ ……(11)

For a liquid

Heat supplied = heat gained by the liquid + heat gained by the calorimeter

IVt = M₁C₁Δθ + M_cC_cΔθ ……(12)

Worked example on specific heat capacity

250 g of lead at 170°C is dropped into 100 g of water at 0°C. If the final steady temperature is 12°C, calculate the specific heat capacity of lead. (C_w = 4.2 × 10³ J kg^-1 K^-1)

Solution

Heat lost by lead = heat gained by water

0.25 × c × (170 – 12) = 0.1 × 4200 × (12 – 0)

c = (420 × 12) ÷ (0.25 × 158)

c = 127.6 J/kg K

Evaluation

1. State the three quantities needed to find the quantity of heat gained or lost by a body.
2. State the unit of heat capacity and the unit of specific heat capacity.
3. In the method of mixtures, which two quantities of heat are equal?
4. Write the expression for the electrical energy supplied to a heating coil in terms of V, R and t.

Assignment

1. Define heat capacity.
2. A metal of mass 0.5 kg is heated to 100°C and transferred to a well-lagged calorimeter of heat capacity 80 J/K containing water of heat capacity 420 J/K at 15°C. If the final steady temperature of the mixture is 25°C, find the specific heat capacity of the metal.
3. Explain the meaning of the statement that the specific heat capacity of a substance is 777 J kg^-1 K^-1.