Specific Latent Heat

$$l = \dfrac{L}{m}$$

where $l$ is the specific latent heat of the substance (in J kg^-1),
$L$ is the heat transfer or energy absorbed, also known as latent heat (in J), and
$m$ is the mass of the substance (in kg)

Specific latent heat is an intrinsic property of the material and is independent of the amount of substance. Different substances have different specific latent heats, which reflect how much energy is required to change their state.

The following is a table of some of the commonly known specific latent heats.

You will notice that the specific latent heat of vaporisation is typically much higher than the specific latent heat of fusion. This is because the process of vaporisation requires breaking more intermolecular bonds compared to fusion.

During fusion (melting), a substance transitions from a solid to a liquid. In this process, the molecules need to overcome the forces that hold them in a fixed position in the solid state, but they are not completely separated from each other. The molecules in the liquid state are still close together, though they can move around more freely.

During vaporisation, the substance transitions from a liquid to a gas. This process requires breaking almost all the intermolecular forces that hold the molecules together in the liquid state. In the gaseous state, the molecules are far apart and move independently, which requires significantly more energy.

Worked Examples

Use $L = ml$ for a change of state at constant temperature. Here $L$ is the energy transferred and $l$ is the specific latent heat. Use $l_f$ for fusion (melting or freezing) and $l_v$ for vaporisation or condensation. If the temperature also changes, calculate that stage separately using $Q = mc\Delta T$.

Example 1: Melting ice at 0 °C

Question: How much energy is needed to melt 150 g of ice at 0 °C into water at 0 °C? Take the specific latent heat of fusion of ice as $334\,000\text{ J kg}^{-1}$.

Step 1: Convert the mass to kilograms: $m = 0.150\text{ kg}$.

Step 2: Use the latent heat equation.

$$L = ml_f = 0.150 \times 334\,000 = 50\,100\text{ J}$$

Answer: 50.1 kJ. The temperature stays at 0 °C during melting; the energy changes the state rather than raising the temperature.

Example 2: Mass of water boiled away

Question: A 500 W heater supplies energy to water already boiling at 100 °C for 3.0 min. Assuming all its energy vaporises water and enough water remains throughout, calculate the mass boiled away. Take $l_v = 2.26 \times 10^6\text{ J kg}^{-1}$.

Step 1: Convert time to seconds and calculate the energy supplied.

$$t = 3.0 \times 60 = 180\text{ s}$$

$$L = Pt = 500 \times 180 = 90\,000\text{ J}$$

Step 2: Rearrange $L = ml_v$.

$$m = \frac{L}{l_v} = \frac{90\,000}{2.26 \times 10^6} \approx 0.0398\text{ kg}$$

Answer: About 0.040 kg (40 g). No energy is needed to bring the water to boiling because it is already at its boiling point.

Example 3: Measuring specific latent heat

Question: A 60 W heater melts 0.024 kg of a solid at its melting point in 120 s. Heat is lost to the surroundings at a constant rate of 10 W. Find the specific latent heat of fusion. Assume the remaining energy is used only for melting.

Step 1: Subtract the power lost from the heater power.

$$P_{\text{useful}} = 60 - 10 = 50\text{ W}$$

Step 2: Find the energy used for melting.

$$L = P_{\text{useful}}t = 50 \times 120 = 6000\text{ J}$$

Step 3: Divide by the mass melted.

$$l_f = \frac{L}{m} = \frac{6000}{0.024} = 250\,000\text{ J kg}^{-1}$$

Answer: $2.5 \times 10^5\text{ J kg}^{-1}$. Ignoring the heat loss would overestimate the specific latent heat because it would count energy that did not melt the solid.

Example 4: Warming ice, melting it and heating the water

Question: Calculate the energy needed to turn 0.10 kg of ice at −10 °C into water at 20 °C. Use $c_{\text{ice}} = 2100\text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1}$, $l_f = 334\,000\text{ J kg}^{-1}$ and $c_{\text{water}} = 4200\text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1}$. Ignore energy absorbed by the container and heat loss to the surroundings.

Stage 1: Warm the ice from −10 °C to 0 °C.

$$Q_1 = mc_{\text{ice}}\Delta T = 0.10 \times 2100 \times [0 - (-10)] = 2100\text{ J}$$

Stage 2: Melt the ice at 0 °C.

$$L = ml_f = 0.10 \times 334\,000 = 33\,400\text{ J}$$

Stage 3: Warm the water from 0 °C to 20 °C.

$$Q_2 = mc_{\text{water}}\Delta T = 0.10 \times 4200 \times 20 = 8400\text{ J}$$

Step 4: Add the energy for all three stages.

$$E_{\text{total}} = 2100 + 33\,400 + 8400 = 43\,900\text{ J}$$

Answer: 43.9 kJ. Use separate specific heat capacities for ice and water, and include the energy needed to melt the ice even though its temperature does not change during that stage.

During freezing or condensation, energy is released. For example, freezing the same 0.10 kg of water at 0 °C releases 33.4 kJ, equal to the energy needed to melt it at 0 °C.

Review the specific heat capacity worked examples for temperature changes without a change of state.