Specific heat capacity (c)
Thermal Physics - OCR A-Level Physics
Key Definition
Specific heat capacity (c)
The energy required to raise the temperature of 1 kg of a substance by 1 K (or $1 ^\circ \text{C}$). Units: $\text{J kg}^{-1} \text{ K}^{-1}$.
The energy required to raise the temperature of 1 kg of a substance by 1 K (or $1 ^\circ \text{C}$). Units: $\text{J kg}^{-1} \text{ K}^{-1}$.
Key Definition
Specific latent heat (L)
The energy required to change the state of 1 kg of a substance without changing its temperature. Units: $\text{J kg}^{-1}$.
The energy required to change the state of 1 kg of a substance without changing its temperature. Units: $\text{J kg}^{-1}$.
$$Q = mc\Delta\theta$$
$$Q = mL$$
- Specific latent heat of fusionThe energy required to change 1 kg of a substance from solid to liquid at its melting point, without changing temperature. ($L_f$): energyThe capacity to do work. Measured in joules (J). to change 1 kg from solid to liquid at the melting point.
- Specific latent heat of vaporisationThe energy required to change 1 kg of a substance from liquid to gas at its boiling point, without changing temperature. ($L_v$): energyThe capacity to do work. Measured in joules (J). to change 1 kg from liquid to gas at the boiling point.
- $L_v$ is always greater than $L_f$ because molecules must completely overcome intermolecular forces to become a gas.
- In experiments, energyThe capacity to do work. Measured in joules (J). losses to the surroundings mean the measured value of c or L is often an overestimate.
Worked Example
A 0.50 kg block of aluminium at $20 ^\circ \text{C}$ is heated to $80 ^\circ \text{C}$. Calculate the energy transferred. ($c_{Al} = 900 \text{ J kg}^{-1} \text{ K}^{-1}$)
Show Solution
1
Identify
$m = 0.50 \text{ kg}$, $c = 900 \text{ J kg}^{-1} \text{ K}^{-1}$, $\Delta\theta = 80 - 20 = 60 \text{ K}$.
2
Apply $Q = mc\Delta\theta$.
3
$Q = 0.50 \times 900 \times 60 = 27000 \text{ J}$.
4
$Q = 27 \text{ kJ}$.
Answer
$Q = 27 \text{ kJ}$
Common Mistake
MEDIUM
Wrong: Using $\Delta\theta$ in $^\circ \text{C}$ for gas law equations but K for specific heat capacityThe energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C)..
Right: For $Q = mc\Delta\theta$, a change of 1 K equals a change of $1 ^\circ \text{C}$, so either unit works for $\Delta\theta$. But gas law equations ($pV = nRT$) always require absolute temperature in K.
Right: For $Q = mc\Delta\theta$, a change of 1 K equals a change of $1 ^\circ \text{C}$, so either unit works for $\Delta\theta$. But gas law equations ($pV = nRT$) always require absolute temperature in K.