Specific Heat Capacity

$$c = \dfrac{Q}{m \Delta T}$$

where $c$ is the specific heat capacity of the substance (in J kg^-1 $\degree$C^-1),
$Q$ is the heat transfer or energy absorbed (in J),
$m$ is the mass of the substance (in kg), and
$\Delta T$ is the change in temperature (in $\degree$C or K)

It is an intrinsic property of the material and is independent of the amount of substance. Different substances have different specific heat capacities, which reflect how much energy is required to change their temperatures. The following is a table of some of the commonly known specific heat capacities.

To calculate the amount of thermal transfer $Q$ needed to raise a substance of specific heat capacity $c$ and mass $m$ by a temperature change of $\Delta T$, we can use this formula: $$Q = mc\Delta T$$

Heat capacity, $C$, is the amount of energy needed to raise the temperature of an object by 1 $\degree$C or 1 K. Unlike specific heat capacity, heat capacity depends on the mass of the object. For an object of mass $m$ made of a substance with specific heat capacity $c$, its heat capacity is given by $$C = mc$$

Worked Examples

Use $Q = mc\Delta T$ when the temperature changes without a change of state. Convert mass to kilograms and use the temperature change, not the final temperature. The values of specific heat capacity needed are given in each question.

Example 1: Energy needed to heat water

Question: Calculate the energy needed to heat 250 g of water from 20 °C to 60 °C. Take the specific heat capacity of water as 4200 J kg⁻¹ °C⁻¹. Ignore energy absorbed by the container and heat loss to the surroundings.

Step 1: Convert the mass and find the temperature rise.

$$m = 0.250\text{ kg}, \qquad \Delta T = 60 - 20 = 40\,{}^{\circ}\text{C}$$

Step 2: Substitute into the equation.

$$Q = mc\Delta T = 0.250 \times 4200 \times 40 = 42\,000\text{ J}$$

Answer: The water needs 42 kJ of energy.

Example 2: Finding specific heat capacity

Question: A 0.50 kg metal block absorbs 9000 J of energy. Its temperature rises from 25 °C to 65 °C. Find its specific heat capacity and its heat capacity. There is no change of state.

Step 1: Find the temperature rise and rearrange the equation.

$$\Delta T = 65 - 25 = 40\,{}^{\circ}\text{C}$$

$$c = \frac{Q}{m\Delta T} = \frac{9000}{0.50 \times 40} = 450\text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1}$$

Step 2: Calculate the heat capacity of the whole block.

$$C = mc = 0.50 \times 450 = 225\text{ J}\,{}^{\circ}\text{C}^{-1}$$

Answer: The material's specific heat capacity is 450 J kg⁻¹ °C⁻¹. The block's heat capacity is 225 J °C⁻¹: it needs 225 J for each 1 °C rise.

Example 3: Heating time with energy losses

Question: A 1000 W heater warms 0.50 kg of water from 20 °C to 80 °C. Only 80% of the electrical energy supplied heats the water. Calculate the heating time. Take $c = 4200\text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1}$.

Step 1: Calculate the energy absorbed by the water.

$$Q = 0.50 \times 4200 \times (80 - 20) = 126\,000\text{ J}$$

Step 2: Find the useful heating power.

$$P_{\text{useful}} = 0.80 \times 1000 = 800\text{ W}$$

Step 3: Use $Q = P_{\text{useful}}t$.

$$t = \frac{126\,000}{800} = 157.5\text{ s} \approx 158\text{ s}$$

Answer: About 158 s (2.63 min). Energy losses increase the time needed to reach the target temperature.

Example 4: Final temperature when water is mixed

Question: Mix 0.20 kg of water at 80 °C with 0.30 kg of water at 20 °C. Find the final temperature, assuming no heat loss to the surroundings, negligible heat capacity of the container and no evaporation.

Step 1: Let the final temperature be $T$. Energy lost by the hot water equals energy gained by the cold water.

$$0.20c(80 - T) = 0.30c(T - 20)$$

Step 2: Both samples are water, so their common specific heat capacity cancels.

$$16 - 0.20T = 0.30T - 6$$

$$22 = 0.50T \qquad \Rightarrow \qquad T = 44\,{}^{\circ}\text{C}$$

Answer: 44 °C. This lies between the starting temperatures and is closer to 20 °C because there is more cold water than hot water.

For calculations involving melting or boiling, see the specific latent heat worked examples.