Heat Capacity

The heat capacity of a substance depends on three factors:

1. Mass of the substance
2. Temperature change (ΔT)
3. Specific heat capacity of the substance

The formula for heat capacity (H.C) is:

Heat Capacity (H.C) = Mass (M) × Specific Heat Capacity (C)

Where:

• M is the mass of the substance (in kg)
• C is the specific heat capacity of the substance (in J/kg°C)

Example 1: Calculating heat capacity

Find the heat capacity of a lump of copper with a mass of 50 kg. The specific heat capacity of copper is 420 J/kg°C.

Given:

Mass of copper, M = 50 kg

Specific heat capacity of copper, C = 420 J/kg°C

Required: Calculate heat capacity (H.C).

Using the formula H.C = M × C:

$H.C = 50 \, \text{kg} \times 420 \, \text{J/kg°C} = 21600 \, \text{J} = 21 \, \text{kJ}$

The quantity of heat (H) required to change the temperature of a body of mass M (in kg) by a temperature change of Q (in °C) is given by:

$H = M \times C \times Q$

Where:

H is the heat gained or lost (in joules)

M is the mass of the body (in kg)

C is the specific heat capacity (in J/kg°C)

Q is the change in temperature (in °C)

Example 2: Calculating the amount of heat supplied to water

Water of mass 3 kg is heated from 26°C to 96°C. Find the amount of heat supplied to the water. The specific heat capacity of water is 4.2 × 10³ J/kg°C.

Given:

Mass of water, M = 3 kg

Specific heat capacity of water, C = 4.2 × 10³ J/kg°C

Initial temperature, $T_i$ = 26°C

Final temperature, $T_f$ = 96°C

Required: Calculate the amount of heat (H).

Using the formula H = M × C × (Tf - Ti):

$H = 3 \, \text{kg} \times 4.2 \times 10^3 \, \text{J/kg°C} \times (96°C - 26°C)$

$H = 3 \times 4.2 \times 10^3 \times 70 = 882000 \, \text{J} = 882 \, \text{kJ}$

Specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius.

The formula for calculating heat using specific heat capacity is:

$H = M \times C \times \Delta T$

Where:

M is the mass (in kg)

C is the specific heat capacity (in J/kg°C)

ΔT is the temperature change (in °C)

To determine the specific heat capacity experimentally, a calorimeter is used. A calorimeter is a specially designed vessel, typically made of polished metal (such as copper or aluminum), with insulating covers and two holes for inserting a thermometer and stirrer.

The heat lost by a solid (Hs) can be calculated by:

$Hs = Ms \times Cs \times (Qs - Qf)$

The heat gained by the calorimeter and stirrer (Hc) is:

$Hc = Mc \times Cc \times (Qf - Qi)$

The heat gained by the water (Hw) is:

$Hw = Mw \times Cw \times (Qf - Qi)$

According to the law of conservation of energy, the heat lost by the solid is equal to the heat gained by the calorimeter and stirrer plus the heat gained by the water:

$Hs = Hc + Hw$

This leads to the equation:

$Ms \times Cs \times (Qs - Qf) = Mc \times Cc \times (Qf - Qi) + Mw \times Cw \times (Qf - Qi)$

Example 3: Finding the final temperature of a system

A piece of metal with a mass of 200 g at a temperature of 100°C is quickly transferred into 50 g of water at 20°C. Find the final temperature of the system, given that the specific heat capacity of water (Cw) is 4200 J/kg°C and the specific heat capacity of the metal (Cm) is 400 J/kg°C.

Using the heat equation, we know:

$\text{Heat gained by water} = 0.5 \, \text{kg} \times 4200 \, \text{J/kg°C} \times (Q - 20)$

$\text{Heat lost by metal} = 0.2 \, \text{kg} \times 400 \, \text{J/kg°C} \times (100 - Q)$

Since no heat is lost to the surroundings, we can set the heat gained by water equal to the heat lost by the metal:

$210 \times (Q - 20) = 80 \times (100 - Q)$

Simplifying the equation:

$21(Q - 20) = 8(100 - Q)$

$21Q - 420 = 800 - 8Q$

$29Q = 1220$

$Q = \frac{1220}{29} = 42.1°C$

Thus, the final temperature of the system is 42.1°C.