Specific Heat Capacity

Heating an object transfers energy to its thermal energy store, raising its temperature. Different materials need different amounts of energy to warm up by the same amount. This property is the specific heat capacity.

The specific heat capacity of a substance is the amount of energy needed to raise the temperature of 1 kg of the substance by 1°C. Its unit is J/kg°C.

A high specific heat capacity means a substance needs a lot of energy to change its temperature. Water has one of the highest values of common materials, about 4200 J/kg°C, which is why the sea warms and cools slowly and why water is used in central-heating systems and car radiators.

Substance	Specific heat capacity (J/kg°C)
Water	4200
Aluminium	900
Copper	385
Lead	130

The table shows why 1 kg of water needs over ten times as much energy to warm by 1°C as 1 kg of copper does.

The energy needed to change the temperature of a substance is found with:

$\Delta E=mc\Delta \theta$

where $\Delta E$ is the change in thermal energy in joules (J), $m$ is the mass in kilograms (kg), $c$ is the specific heat capacity in J/kg°C, and $\Delta \theta$ is the temperature change in °C. The symbol $\theta$ (theta) stands for temperature, and $\Delta$ means "change in".

This equation is given on the Physics equation sheet, so you do not have to memorise it. You do need to rearrange and apply it confidently.

The temperature change $\Delta \theta$ is the difference between the final and starting temperatures. Because it is a change, a rise from 15°C to 45°C is a $\Delta \theta$ of 30°C, whether you measure in °C or kelvin, since a change of 1°C equals a change of 1 K.

Worked example — how much energy is needed to raise the temperature of 2 kg of water from 20°C to 50°C? Take $c=4200$ J/kg°C.

First find the temperature change:

$\Delta \theta =50-20=30\text{°C}$

Then substitute into the equation:

$\Delta E=mc\Delta \theta =2\times 4200\times 30=252000\text{J}=252\text{kJ}$

So 252 000 J are transferred to the water's thermal store.

If instead you were told 252 000 J were supplied and asked for the temperature rise, you would rearrange to $\Delta \theta =\frac{\Delta E}{mc}$:

$\Delta \theta =\frac{252000}{2\times 4200}=\frac{252000}{8400}=30\text{°C}$

which returns the same figure, confirming the rearrangement.

Required practical 1 determines the specific heat capacity of a solid, usually a metal block. The aim is to link the energy input to the rise in temperature and thermal energy.

Apparatus

• A metal block of known mass (for example a 1 kg aluminium block) with holes for a heater and a thermometer
• A 12 V electric immersion heater
• A joulemeter (or an ammeter, voltmeter and stopwatch to calculate energy)
• A thermometer and insulation to wrap around the block

Method

1. Measure the mass $m$ of the block on a balance.
2. Insert the heater and thermometer into the block and record the starting temperature.
3. Wrap the block in insulation to reduce energy transfer to the surroundings.
4. Switch on the heater and start timing. Record the energy supplied (from the joulemeter, or from $E=VIt$) and the temperature at regular intervals.
5. Continue until a clear temperature rise is measured, then find the total energy input $\Delta E$ and the total temperature change $\Delta \theta$.

Each step of the practical controls a source of error and gives you the data the equation needs.

Variable	Role	Reason
Energy input	Independent variable	It is what you control and measure as it is supplied
Temperature	Dependent variable	It responds to the energy supplied
Mass, material	Control variables	Kept the same so the value of $c$ found is for that one substance

The insulation reduces energy transferred to the surroundings, so more of the energy input actually heats the block. A little oil in the thermometer hole improves thermal contact so the reading tracks the block's true temperature. You wait before reading because energy takes time to spread evenly through the block.

Rearrange the equation to find $c$:

$c=\frac{\Delta E}{m\Delta \theta }$

Worked example — the heater supplies 18 000 J to a 1 kg aluminium block, whose temperature rises by 20°C. Find $c$.

$c=\frac{18000}{1\times 20}=900\text{J/kg°C}$

This matches the accepted value for aluminium. Measured values are often slightly high because some energy is lost to the surroundings, so less energy than counted actually warmed the block.

1. Forgetting that Δθ is a change, not a temperature

Use the difference between final and starting temperatures. A jump from 20°C to 50°C gives $\Delta \theta =30$, not 50.

2. Leaving mass in grams

Mass in $\Delta E=mc\Delta \theta$ must be in kilograms. A 500 g sample is 0.5 kg.

3. Confusing specific heat capacity with specific latent heat

Specific heat capacity applies while the temperature is changing. Specific latent heat applies during a change of state, when the temperature stays constant. They are different quantities with different equations.

4. Explaining a high measured value wrongly

A measured $c$ is usually a little too high because energy escapes to the surroundings, so the recorded temperature rise is smaller than it would be for the counted energy. Insulation reduces this error.

5. Not stating the units of c

Specific heat capacity is measured in J/kg°C. Dropping the units or quoting J/kg loses the mark.