Heating, Heat Capacity and Latent Heat

Heating changes a system's internal energy

A system is the object or group of objects we choose to study—for example, the water inside a kettle. Its internal energy is the total kinetic and potential energy of its particles: energy associated with their movement and their positions relative to one another.

Energy can be transferred to a material by heating it with a fuel or an electric heater. Doing work on a material can also increase its internal energy: rubbing your hands together warms them without a flame or heater. Heating and work are different ways of transferring the same conserved quantity, energy.

An increase in internal energy can raise a material's temperature or cause a change of state. These changes need different equations.

Specific heat capacity: energy for a temperature change

Specific heat capacity, $c$, is the energy required to raise the temperature of 1 kg of a substance by 1°C. Its unit is J/kg/°C.

A substance with a larger specific heat capacity needs more energy for the same mass and temperature rise. Equivalently, if equal masses receive equal amounts of energy, the material with the larger specific heat capacity has the smaller temperature rise.

For a temperature change without a change of state:

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

Here $\Delta E$ is the change in internal energy in joules (J), $m$ is mass in kilograms (kg), and $\Delta\theta$ is temperature change in degrees Celsius (°C). Calculate the temperature change using:

$$\Delta\theta = \theta_{\text{final}}-\theta_{\text{initial}}$$

For example, suppose 0.50 kg of water, with $c = 4200\text{ J/kg/°C}$, warms from 20°C to 40°C. Its temperature rise is 20°C, so its internal energy increases by $0.50\times4200\times20 = 42\,000\text{ J}$. Twice as much water would need twice as much energy for that rise.

When a material cools, its internal energy decreases and energy is transferred to the surroundings. The same relationship gives the magnitude of the energy released if you use the size of the temperature fall.

Specific latent heat: energy for a change of state

Specific latent heat, $L$, is the energy required to change the state of 1 kg of a substance without changing its temperature. Its unit is J/kg: unlike specific heat capacity, it has no temperature-change factor.

During melting or boiling, energy continues to enter the substance even though its temperature remains constant. Its internal energy increases as the particles' arrangement and separation change, rather than through a temperature rise.

There are two commonly used forms:

• Specific latent heat of fusion applies to melting and freezing.
• Specific latent heat of vaporisation applies to boiling and condensation.

Melting and vaporisation require an input of energy. Freezing and condensation release energy.

The energy needed for a change of state is:

$$E = mL$$

Here $E$ is energy in J, $m$ is the mass that changes state in kg, and $L$ is specific latent heat in J/kg.

Suppose 0.20 kg of a solid is already at its melting temperature and has a given specific latent heat of fusion of 150 000 J/kg. Melting it requires $0.20\times150\,000 = 30\,000\text{ J}$. That energy melts the material; it does not raise its temperature.

If a solid starts below its melting temperature, first calculate the energy to warm it using $mc\Delta\theta$, then the energy to melt it using $mL$. Add these separate energy transfers to find the total.

Following energy through a kettle, brakes and a collision

In an electric kettle, energy is transferred electrically from the supply to the heating element, then by heating to the water. The internal energy of the water increases and its temperature rises. The kettle itself and the surrounding air also warm, so not all the supplied energy reaches the water.

Once the water is boiling, further energy transfers cause liquid water to become gas rather than raising the temperature of the boiling water. Reaching boiling temperature and changing all the water into gas are therefore different calculations.

When an object slows down, its kinetic energy store decreases. During braking, friction transfers energy to the internal energy stores of the brakes and other parts of the surroundings, increasing their temperatures. Air resistance can also transfer energy to the surroundings.

When a moving object hits an obstacle, its kinetic energy decreases. Energy may temporarily be stored elastically as the objects deform. Permanent deformation and friction increase the internal energy of the objects, while sound transfers some energy to the surroundings. Energy is conserved, but it becomes spread among more stores.

Linking electrical and mechanical calculations to heating

For an electric heater, calculate the supplied energy using:

$$E = Pt = VIt$$

Power $P$ is measured in watts (W), potential difference $V$ in volts (V), current $I$ in amperes (A), and time $t$ in seconds (s).

For example, a 100 W heater operating for 60 s supplies 6000 J. If all this energy heats a 0.50 kg block with $c = 400\text{ J/kg/°C}$, its temperature rise is:

$$\Delta\theta = \frac{E}{mc} = \frac{6000}{0.50\times400} = 30\text{°C}$$

For a moving object, its kinetic energy is:

$$E_k = \frac{1}{2}mv^2$$

Here $m$ is the moving object's mass in kg and $v$ is its speed in m/s. Calculate the decrease in kinetic energy, then use that energy in $mc\Delta\theta$ if it heats a material.

For example, a 2.0 kg object moving at 10 m/s has 100 J of kinetic energy. If it stops and all 100 J heats a 0.10 kg part with $c = 500\text{ J/kg/°C}$, that part warms by $100\div(0.10\times500) = 2.0\text{°C}$. If some energy heats other objects or is transferred by sound, its temperature rise will be smaller.

Measuring specific heat capacity safely

A metal block can be heated electrically to measure its specific heat capacity. You need a balance, a block with holes for an electric heater and thermometer, suitable insulation, a low-voltage d.c. supply, an ammeter, a voltmeter and a stopwatch.

1. Measure the block's mass in kg. Fit the heater and thermometer into their holes with good thermal contact, and wrap the block in suitable insulation to reduce energy transfer to the surroundings.
2. With the power off, connect the ammeter in series with the heater and the voltmeter in parallel across it.
3. Record the initial temperature. Switch on the heater and start timing together.
4. Record the current and potential difference, checking that they remain steady. Take temperature readings at regular intervals.
5. Once there is a clear temperature rise, record the final temperature and elapsed time, then switch off. A modest rise of around 10°C can provide a useful measurement; there is no need to make the block extremely hot.
6. Calculate the electrical energy supplied using $E = VIt$, then calculate:

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

Use the low-voltage supply rather than exposed mains connections. Keep leads secure, avoid touching the hot block or heater, and allow them to cool before handling or repeating the experiment. Do not switch on a heater outside the block, and avoid overheating insulation that could soften or be damaged.

The calculation assumes that all the supplied energy increases the block's internal energy. In practice, some heats the heater and thermometer or escapes to the surroundings. This makes the temperature rise smaller and can make the calculated specific heat capacity too high. Insulation reduces this problem. Repeating the measurement after cooling helps check consistency.

For a liquid, the same principle can be used with a measured mass in an insulated container, a lid, an immersed heater and a thermometer. Keep electrical connections dry and handle hot liquid carefully.

Temperature change or state change?

• Temperature change: $\Delta E = mc\Delta\theta$, with $c$ in J/kg/°C.
• State change at constant temperature: $E = mL$, with in J/kg.