3.6.2.1
Specific heat capacity: the energy to raise 1 kg by 1 degree
Thermal Energy Transfer — AQA A-Level Physics
Key Definition
Specific heat capacity — The amount of thermal energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 degree C) without a change of state. Symbol: c. Unit: J kg^{-1} \(K^{-1}\).
$$Q = mc\Delta\theta$$
$$\Delta Q = mc\Delta\theta$$
- $\Delta Q$: change in thermal energyThe capacity to do work. Measured in joules (J). (J)
- $m$: mass of substance (kg)
- $c$: specific heat capacityThe energyThe capacity to do work. Measured in joules (J). required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). (J kg^{-1} \(K^{-1}\))
- $\Delta \theta$: change in temperature (K or degrees C)
- A low specific heat capacityThe energyThe capacity to do work. Measured in joules (J). required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). means the substance heats up and cools down quickly (e.g. copper, 390 J kg^{-1} \(K^{-1}\)).
- A high specific heat capacityThe energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). means it heats up and cools down slowly (e.g. water, 4200 J kg^{-1} \(K^{-1}\)).
- The temperature change delta theta is the same in kelvin and Celsius. No conversion needed.
Worked Example
A 1.7 kW kettle heats 650 g of liquid from 25 degrees C to boiling (100 degrees C) in 3.5 minutes. Calculate the specific heat capacity.
Show Solution
1
Calculate energy supplied
$$E = Pt = 1700 \times (3.5 \times 60) = 1700 \times 210 = 3.57 \times 10^5 \text{ J}$$
2
Find temperature change
$$\Delta\theta = 100 - 25 = 75 \text{ degrees C}$$
3
Rearrange and substitute
$$c = \frac{\Delta Q}{m\Delta\theta} = \frac{3.57 \times 10^5}{0.650 \times 75} = 7300 \text{ J kg}^{-1} \text{ degrees C}^{-1}$$
Answer
$c = 7300$ J kg$^{-1}$ degrees C$^{-1}$
Examiner Tips and Tricks
- The temperature difference is the same in kelvin and Celsius, so there is no need to convert.
- This is because a 1 K change equals a 1 degree C change.