Specific Heat Capacity

Specific heat capacity formula

The energy required to increase the temperature of a substance by a certain amount depends on two factors:

• The mass - the amount of a substance there is. The greater the mass, the more energy will be required to heat it up.
• The material - the temperature of different materials will increase by different amounts when energy is applied to them.

The amount that a material heats up when energy is applied to it depends on its specific heat capacity, \( c \). The greater a material's specific heat capacity, the more energy is required for its temperature to increase by a given amount. The specific heat capacities of various materials are shown in the table below.

The table shows that non-metals generally have a higher specific heat capacity than metals. Also, water has a very high specific heat capacity compared to other materials. It's value is \( 4200\,\mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1} \), meaning that \( 4200\,\mathrm J \) of energy is required to heat up \( 1 \,\mathrm kg \) of water by \( 1\,\mathrm K \). It takes a lot of energy to heat up water and, on the other hand, water takes a long time to cool down.

Now that we have discussed what factors affect how the temperature of a substance changes, we can state the specific heat capacity formula. The change in energy, \( \Delta E \), required to produce a certain change in temperature, \( \Delta\theta \), in a material of mass \( m \) and specific heat capacity \( c \) is given by the equation

$\Delta E=mc\Delta\theta,$

which in words can be written as

$\text{change}\;\text{in}\;\text{energy}=\text{mass}\times \text{specific}\;\text{heat}\;\text{capacity}\times \text{change}\;\text{in}\;\text{temp}.$

Notice that this equation relates the change in energy to the change in temperature. The temperature of a substance decreases when energy is taken away from it, in which case the quantities \( \Delta E \) and \( \Delta\theta \) will be negative.

SI unit of specific heat capacity

As you may have noticed from the table in the section above, the SI unit for specific heat capacity is \( \mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1} \). It can be derived from the specific heat capacity equation. Let us first rearrange the equation to find an expression for the specific heat capacity on its own:

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

The SI units for the quantities in the equation are as follows:

• Joules \( \mathrm J \), for energy.
• Kilograms \( \mathrm{kg} \), for mass.
• Kelvin \( \mathrm K \), for temperature.

We can plug the units into the equation for specific heat capacity to find the SI unit for \( c \):

$unit(c)=\frac{\mathrm J}{\mathrm{kg}\,\mathrm K}=\mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1}.$

As we are only dealing with a change in temperature - a difference between two temperatures rather than a single temperature - the units can be either Kelvin, \( \mathrm K \), or degrees Celsius, \( ^\circ \mathrm C \). The Kelvin and Celsius scales have the same divisions and only differ in their starting points - \( 1\,\mathrm K \) is equal to \( 1 ^\circ\mathrm C \).

Specific heat capacity method

A short experiment can be performed to find the specific heat capacity of a block of material, such as aluminium. Below is a list of the equipment and materials needed:

• Thermometer.
• Stopwatch.
• Immersion heater.
• Power supply.
• Ammeter.
• Voltmeter.
• Connecting wires.
• Aluminium block of known mass with holes for the thermometer and the immersion heater to be placed in.

This experiment uses an immersion heater to increase the temperature of an aluminium block so that the specific heat capacity of aluminium can be measured. The setup is shown in the image below. First, the immersion heater circuit needs to be constructed. The immersion heater should be connected to a power supply in series with an ammeter and placed in parallel with a voltmeter. Next, the heater can be placed inside the corresponding hole in the block and the same should be done for the thermometer.

Once everything is set up, turn on the power supply and start the stopwatch. Note the initial temperature of the thermometer. Take readings of the current from the ammeter and the voltage from the voltmeter every minute for a total of \( 10 \) minutes. When the time is up, note the final temperature.

In order to calculate the specific heat capacity, we must find the energy transferred to the block by the heater. We can use the equation

$E=Pt,$

Once everything is set up, turn on the power supply and start the stopwatch. Note the initial temperature of the thermometer. Take readings of the current from the ammeter and the voltage from the voltmeter every minute for a total of \( 10 \) minutes. When the time is up, note the final temperature.

In order to calculate the specific heat capacity, we must find the energy transferred to the block by the heater. We can use the equation

$E=Pt,$

where \( E \) is the energy transferred in Joules \( \mathrm J \), \( P \) is the power of the immersion heater in Watts \( \mathrm W \), and \( t \) is the heating time in seconds \( \mathrm s \). The power of the heater can be calculated by using

$P=IV,$

where \( I \) is the ammeter current in Amps \( \mathrm A \), and \( V \) is the voltage measured by the voltmeter in volts \( \mathrm V \). You should use your average current and voltage values in this equation. This means that the energy is given by

$E=IVt.$

We already found an equation for specific heat capacity as

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

Now that we have an expression for the energy transferred to the aluminium block, we can substitute this into the specific heat capacity equation to get

$c=\frac{IVt}{m\Delta\theta}.$

After completing this experiment, you will have all of the quantities needed to calculate the specific heat capacity of aluminium. This experiment can be repeated to find the specific heat capacities of different materials.

There are several sources of error in this experiment that should be avoided or noted:

• The ammeter and voltmeter must both be initially set to zero so that the readings are correct.
• A small amount of energy is dissipated as heat in the wires.
• Some energy supplied by the immersion heater will be wasted - it will heat up the surroundings, the thermometer, and the block. This will result in the measured specific heat capacity being less than the true value. The proportion of wasted energy can be reduced by insulating the block.
• The thermometer must be read at eye level to record the correct temperature.

Specific heat capacity calculation

The equations discussed in this article can be used for many practice questions about specific heat capacity.