Mastering Specific Heat Capacity for Your Physics Exam
Demystify specific heat capacity with this targeted revision guide. Learn what examiners look for, tackle worked examples, avoid common pitfalls, and ace your exam.
What the examiner is testing
Examiners are testing your ability to apply the specific heat capacity formula to calculate energy changes, temperature changes, or specific heat capacity itself, often in practical contexts. Marks are typically awarded for correctly identifying and substituting values into the formula, and for the final answer with correct units.
The method
- Identify the substance undergoing a temperature change.
- Determine the mass of the substance, \(m\), in kilograms (kg). Convert from grams if necessary.
- Identify the specific heat capacity, \(c\), of the substance in joules per kilogram per degree Celsius (\(\text{J kg}^{-1} ^\circ\text{C}^{-1}\)) or joules per kilogram per Kelvin (\(\text{J kg}^{-1} \text{K}^{-1}\)).
- Find the initial and final temperatures of the substance to calculate the temperature change, \(\Delta\theta\), in degrees Celsius (\(^\circ\text{C}\)) or Kelvin (K).
- State the specific heat capacity formula: \(E = mc\Delta\theta\).
- Rearrange the formula if you are calculating \(m\), \(c\), or \(\Delta\theta\).
- Substitute the known values into the formula.
- Calculate the unknown quantity, ensuring your answer includes the correct units.
Worked example
A 0.5 kg block of aluminium is heated, causing its temperature to rise from \(20^\circ\text{C}\) to \(70^\circ\text{C}\). The specific heat capacity of aluminium is \(900 \text{ J kg}^{-1} ^\circ\text{C}^{-1}\). Calculate the energy transferred to the aluminium block.
$$
\begin{aligned}
m &= 0.5 \text{ kg} \\
c &= 900 \text{ J kg}^{-1} ^\circ\text{C}^{-1} \\
\Delta\theta &= 70^\circ\text{C} - 20^\circ\text{C} = 50^\circ\text{C} \\
E &= mc\Delta\theta \\
E &= (0.5 \text{ kg}) \times (900 \text{ J kg}^{-1} ^\circ\text{C}^{-1}) \times (50^\circ\text{C}) \\
E &= 450 \text{ J kg}^{-1} ^\circ\text{C}^{-1} \text{ kg} \times 50^\circ\text{C} \\
E &= 22500 \text{ J}
\end{aligned}
$$
Sanity check: Aluminium has a moderate specific heat capacity. Heating 0.5 kg by \(50^\circ\text{C}\) requires a significant amount of energy, so 22.5 kJ seems reasonable.
Worked example: a harder one
An immersion heater supplies \(1.5 \text{ kJ}\) of energy to \(200 \text{ g}\) of a liquid, causing its temperature to increase by \(15^\circ\text{C}\). Calculate the specific heat capacity of the liquid.
The obvious first move is to substitute the given values directly into \(E = mc\Delta\theta\). However, the energy is in kilojoules and the mass is in grams, which will lead to incorrect units and an incorrect numerical answer.
$$ \begin{aligned} E &= 1.5 \text{ kJ} = 1.5 \times 1000 \text{ J} = 1500 \text{ J} \\ m &= 200 \text{ g} = 200 \div 1000 \text{ kg} = 0.2 \text{ kg} \\ \Delta\theta &= 15^\circ\text{C} \\ E &= mc\Delta\theta \\ c &= \frac{E}{m\Delta\theta} \\ c &= \frac{1500 \text{ J}}{(0.2 \text{ kg}) \times (15^\circ\text{C})} \\ c &= \frac{1500 \text{ J}}{3 \text{ kg } ^\circ\text{C}} \\ c &= 500 \text{ J kg}^{-1} ^\circ\text{C}^{-1} \end{aligned} $$
Practice
- Calculate the energy required to raise the temperature of \(2 \text{ kg}\) of water by \(10^\circ\text{C}\). (Specific heat capacity of water = \(4200 \text{ J kg}^{-1} ^\circ\text{C}^{-1}\)).
- A metal block of mass \(0.8 \text{ kg}\) absorbs \(12000 \text{ J}\) of energy, and its temperature rises from \(25^\circ\text{C}\) to \(40^\circ\text{C}\). Calculate the specific heat capacity of the metal.
- An electric kettle contains \(1.5 \text{ kg}\) of water. The kettle has a power rating of \(2 \text{ kW}\). How long will it take for the water's temperature to increase by \(80^\circ\text{C}\)? Assume all energy from the kettle is transferred to the water. (Specific heat capacity of water = \(4200 \text{ J kg}^{-1} ^\circ\text{C}^{-1}\)).
- A student investigates the specific heat capacity of a liquid. They heat \(250 \text{ g}\) of the liquid using an immersion heater rated at \(50 \text{ W}\) for \(3 \text{ minutes}\). The temperature of the liquid rises from \(22^\circ\text{C}\) to \(38^\circ\text{C}\).
Calculate the specific heat capacity of the liquid. State any assumptions made.
Answers:
- \(E = (2 \text{ kg}) \times (4200 \text{ J kg}^{-1} ^\circ\text{C}^{-1}) \times (10^\circ\text{C}) = 84000 \text{ J}\)
- \(\Delta\theta = 40^\circ\text{C} - 25^\circ\text{C} = 15^\circ\text{C}\)
\(c = \frac{E}{m\Delta\theta} = \frac{12000 \text{ J}}{(0.8 \text{ kg}) \times (15^\circ\text{C})} = 1000 \text{ J kg}^{-1} ^\circ\text{C}^{-1}\) - \(P = 2 \text{ kW} = 2000 \text{ W}\)
\(E = mc\Delta\theta = (1.5 \text{ kg}) \times (4200 \text{ J kg}^{-1} ^\circ\text{C}^{-1}) \times (80^\circ\text{C}) = 504000 \text{ J}\)
\(E = Pt \implies t = \frac{E}{P} = \frac{504000 \text{ J}}{2000 \text{ W}} = 252 \text{ s}\) - \(m = 250 \text{ g} = 0.25 \text{ kg}\)
\(P = 50 \text{ W}\)
\(t = 3 \text{ minutes} = 3 \times 60 \text{ s} = 180 \text{ s}\)
\(E = Pt = 50 \text{ W} \times 180 \text{ s} = 9000 \text{ J}\)
\(\Delta\theta = 38^\circ\text{C} - 22^\circ\text{C} = 16^\circ\text{C}\)
\(c = \frac{E}{m\Delta\theta} = \frac{9000 \text{ J}}{(0.25 \text{ kg}) \times (16^\circ\text{C})} = \frac{9000 \text{ J}}{4 \text{ kg } ^\circ\text{C}} = 2250 \text{ J kg}^{-1} ^\circ\text{C}^{-1}\)
Assumption: All the energy supplied by the heater is transferred to the liquid (no energy is lost to the surroundings or absorbed by the container).
The three mistakes that lose marks
- Incorrect units for mass: Using mass in grams instead of kilograms. This results in an answer that is 1000 times too large or too small. For example, if \(m = 200 \text{ g}\) is used as \(200\) instead of \(0.2 \text{ kg}\), and \(c\) is calculated, it will be \(1000\) times smaller than the correct value.
- Incorrect units for energy: Using energy in kilojoules instead of joules. This also leads to an answer that is 1000 times too large or too small. For example, if \(E = 1.5 \text{ kJ}\) is used as \(1.5\) instead of \(1500 \text{ J}\), and \(c\) is calculated, it will be \(1000\) times smaller than the correct value.
- Confusing temperature and temperature change: Using the final temperature instead of the change in temperature (\(\Delta\theta\)). For instance, if a substance heats from \(20^\circ\text{C}\) to \(70^\circ\text{C}\), using \(70^\circ\text{C}\) for \(\Delta\theta\) instead of \(50^\circ\text{C}\) will give an incorrect energy value.
30-second recap
Specific heat capacity (\(c\)) is the energy required to raise the temperature of \(1 \text{ kg}\) of a substance by \(1^\circ\text{C}\) (or \(1 \text{ K}\)). The formula \(E = mc\Delta\theta\) relates energy transferred (\(E\)), mass (\(m\)), specific heat capacity (\(c\)), and temperature change (\(\Delta\theta\)). Always ensure consistent units (joules, kilograms, degrees Celsius/Kelvin) when applying the formula.