# Heat Capacity & Specific Heat Capacity

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## Heat Capacity

Heat Capacity, $C$, of a body is defined as the amount of heat ($Q$) required to raise its temperature ($\theta$) by one degree, without going through a change of state.

• Amount of heat needed to raise the temperature of an object depends on the mass of the object.
• Heat capacity also depends on the material of the object. Some materials are harder to heat up than others. The molecules in a liquid such as water require more energy to move faster than copper atoms in a solid. So, in order to record 1°C increase in temperature, liquids would require more heat energy than solids.
• SI. unit of heat capacity is joule per kelvin ($\text{J K}^{-1}$) or joule per degree Celsius ($\text{J }^{\circ}\text{C}^{-1}$).

$C = \frac{Q}{\Delta \theta}$ , where

$C$ = heat capacity ($\text{J K}^{-1}$, $\text{J }^{\circ}\text{C}^{-1}$)

$Q$ = heat or thermal energy absorbed or released ($\text{J}$)

$\Delta \theta$ = change in temperature ($\text{K}$ or $^{\circ}\text{C}$)

## Specific Heat Capacity

Specific heat capacity, $c$, of a body is defined as the amount of heat ($Q$) required to raise the temperature ($\theta$) of a unit mass of it by one degree, without going through a change in state.

• When the mass of an object is greater, the object will contain more atoms or molecules than a less massive object made up of the same material. Hence, when the temperature of the objects are raised, the more massive object will require a larger thermal energy than the less massive object. (Analogy: The more massive object has more “mouths” to feed) It is thus more common to consider the heat capacity per unit mass or specific heat capacity of the body.

SI unit of specific heat capacity is joule per kilogram per kelvin ($\text{J kg}^{-1}\text{ K}^{-1}$) or joule per kilogram per degree Celsius ($\text{J kg}^{-1} \, ^{\circ}\text{C}^{-1}$)

$$Q = mc \Delta \theta$$

, where

$c$ = specific heat capacity ($\text{J kg}^{-1}\text{ K}^{-1}$, $\text{J kg}^{-1} \, ^{\circ}\text{C}^{-1}$)
$m$ = mass of substance ($\text{kg}$)
$Q$ = heat or thermal energy absorbed or released ($\text{J}$)
$\Delta \theta$ = change in temperature ($\text{K}$ or $^{\circ}\text{C}$)

Specific heat capacity of gases is higher than that of liquids and much higher than that of liquids and much higher than that of solids. The substances with higher specific heat capacity cool or warm very slowly compared to substances with lower specific heat capacity.

With gases, the molar heat capacity (the heat capacity of one mole of a gas at constant pressure or constant volume) is generally more useful than the specific heat capacity, which is based on mass.

### Significance Of Water’s Elevated Specific Heat Capacity

Water possesses a specific heat capacity of 4200 $\text{J kg}^{-1} \, ^{\circ}\text{C}^{-1}$, whereas soil has a specific heat capacity of approximately 800 $\text{J kg}^{-1} \, ^{\circ}\text{C}^{-1}$. This distinction results in the ocean’s temperature exhibiting more gradual changes compared to land. To raise the temperature of a given mass by 1°C, water requires five times more thermal energy than an equivalent mass of soil. Similarly, water releases more energy when cooling by 1°C. As islands are surrounded by water, they undergo less pronounced temperature variations between summer and winter compared to expansive land masses like Central Asia. The elevated specific heat capacity of water, coupled with its cost-effectiveness and widespread availability, explains its utilization in engine cooling and central heating system radiators.

### Determining Specific Heat Capacity Of A Liquid

Materials Needed:

1. Liquid sample
2. Calorimeter
3. Thermometer
4. Heater or heat source
5. Stopwatch

Procedure:

1. Select a Liquid Sample:
Choose the liquid for which you want to determine the specific heat capacity. Ensure the liquid is pure and devoid of impurities to obtain accurate results.
2. Measure the Mass and Initial Temperature:
Use a graduated cylinder to measure the mass of the liquid accurately. Record the initial temperature of the liquid using a thermometer.
3. Place the Liquid in the Calorimeter:
Transfer the liquid to the calorimeter, a container designed to measure heat exchanges. Make sure the calorimeter is clean and dry.
4. Heat the Liquid:
Introduce a known amount of heat to the liquid using a heater or another heat source. Monitor the temperature rise within the calorimeter.
$$\text{energy transferred to liquid} = \text{power of heater} \times \text{time heater is on}$$
5. Record Temperature Changes:
Continuously measure and record the temperature changes over time using a thermometer. This data will be essential for subsequent calculations.
6. Calculate Heat Absorbed:
Utilize the recorded temperature changes and the heat capacity of the calorimeter to calculate the heat absorbed by the liquid using the formula:
$$Q = mc \Delta \theta$$
where $Q$ is the heat absorbed, $m$ is the mass of the liquid, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change.
7. Determine Specific Heat Capacity:
Solve for the specific heat capacity $c$ using the rearranged formula:
$$c = \frac{Q}{m \times \Delta T}$$
8. Repeat for Accuracy:
For reliable results, repeat the experiment several times and calculate an average specific heat capacity.

### Determining Specific Heat Capacity Of A Solid

Materials Needed:

1. Solid sample
2. Calorimeter
3. Thermometer
4. Heater or heat source
5. Stopwatch
6. Ruler or calipers

Procedure:

1. Select a Solid Sample:
Choose the solid for which you want to determine the specific heat capacity. Ensure it is homogeneous and free from any irregularities.
2. Measure the Dimensions and Mass:
Measure the dimensions of the solid using a ruler or calipers. Calculate the volume and, if possible, use the density to determine the mass accurately.
3. Record the Initial Temperature:
Record the initial temperature of the solid using a thermometer.
4. Place the Solid in the Calorimeter:
Transfer the solid to the calorimeter, ensuring the calorimeter is clean and dry.
5. Heat the Solid:
Apply a known amount of heat to the solid using a heater or heat source. Monitor the temperature changes within the calorimeter.
6. Record Temperature Changes:
Continuously measure and record the temperature changes over time using a thermometer.
7. Calculate Heat Absorbed:
Use the recorded temperature changes and the heat capacity of the calorimeter to calculate the heat absorbed by the solid, following the same formula as for liquids:
$$Q = mc \Delta \theta$$
8. Determine Specific Heat Capacity:
Solve for the specific heat capacity $c$ using the rearranged formula.
9. Repeat for Accuracy:
Conduct the experiment multiple times to enhance precision, calculating an average specific heat capacity for the solid.

## Worked Examples

### Example 1

In a simple experiment, 100 g of water requires 12 600 J of heat to raise it from 30 °C to 60 °C.

i) Find the heat capacity of 100 g of water.

ii) Find the heat capacity of 1000 g of water.

iii) Find the heat needed to raise 1000 g of water from 30 °C to 40 °C.

i) $C = \frac{Q}{\Delta \theta} = \frac{12600}{30}$

C = 420 JK-1

ii) Since 1000 g of water has 10 times the mass of 100 g of water, the heat capacity of 1000 g of water = 10 times the heat capacity of 100 g of water.

Hence, C for 1000 g of water = 4200 JK-1

iii) Heat needed, $Q = C \Delta \theta = 4200 (40 – 30) = 42000 J$

### Example 2

An electric heating coil supplies 50 W of power to a metal block of mass 0.60 kg and raises the temperature of the block from 20 °C to 45 °C in 90 s. Calculate the specific heat capacity of the metal. What assumption did you make to arrive at your answer?

Assumption: No heat loss to the surroundings, i.e. all the heat supplied by the heater is absorbed by the metal.

Heat energy supplied by heater, E = P x t = 4500 J

Recall: $E = mc \Delta \theta$

$c = \frac{4500}{0.60 \times 25}$

c = 300 Jkg-1°C-1

### Example 3

Determine the specific heat capacity of a 5 kg mass when supplied with 20,000 J of energy, resulting in a temperature increase from 15°C to 25°C.

\begin{aligned} c &= \frac{\Delta E}{m \Delta \theta} \\ &= \frac{20 \, 000 \text{ J}}{5 \text{ kg} \times \left( 25-15 \right)^{\circ}\text{C}} \\ &= \frac{20 \, 000 \text{ J}}{50 \text{ kg }^{\circ}\text{C}} \\ &= 400 \text{ J kg}^{-1} \, ^{\circ}\text{C}^{-1} \end{aligned}