# Specific latent heat

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u20-specific-latent-heat/

This module introduces specific latent heat, the heat energy involved in phase changes at constant temperature. You will learn to solve calculation problems and understand experimental methods for CIE A-Level Physics exams.

**Prerequisites:** [Specific heat capacity](https://www.owlsprep.com/study/cie-9702-u20-specific-heat-capacity/)

## Learning objectives

- Define specific latent heat and distinguish between fusion and vaporization
- Use $Q = mL$ to solve thermal calorimetry problems
- Explain why temperature remains constant during phase change
- Describe experimental methods to measure specific latent heat for CIE practical exams

## Core Definition of Specific Latent Heat

When a substance changes phase (solid to liquid, liquid to gas), heat energy is absorbed or released, but temperature does not change. This is because energy is used to break or form intermolecular bonds, not to change the average kinetic energy of molecules (which determines temperature).

**Specific Latent Heat** — The amount of heat energy required to change the phase of 1 kilogram of a substance at constant temperature.

*Notation:* $L$

*Example:* Water has two common specific latent heat values: one for melting, one for boiling.

$$Q = mL$$

- - **Specific latent heat of fusion ($L_f$)**: For solid ↔ liquid phase change. For water, $L_f = 3.34 \times 10^5$ J kg⁻¹.
- - **Specific latent heat of vaporization ($L_v$)**: For liquid ↔ gas phase change. For water, $L_v = 2.26 \times 10^6$ J kg⁻¹.

**Worked example:** How much heat energy is required to melt 250 g of ice at 0°C? Specific latent heat of fusion of ice is $3.34 \times 10^5$ J kg⁻¹.

1. Convert mass from grams to kilograms:
2. $$m = 250 \text{ g} = 0.25 \text{ kg}$$
3. Substitute into the formula $Q=mL_f$:
4. $$Q = 0.25 \times 3.34 \times 10^5 = 8.35 \times 10^4 \text{ J}$$
5. 83500 J of heat is required to melt the ice.

*Calculator:* allowed

## Solving Calorimetry Problems

Most CIE calculation problems use the principle of energy conservation: for an insulated system, heat lost by a hot substance equals heat gained by the substance undergoing phase change. We combine $Q=mL$ for phase change with $Q=mc\Delta T$ for temperature changes.

> **tip**
>
> Always confirm the direction of energy flow: heat flows from higher temperature to lower temperature. For example, condensing steam releases latent heat to warm cooler surrounding water.

**Worked example:** A 50 g block of ice at 0°C is added to 200 g of water at 40°C in an insulated container. Find the final temperature of the mixture, assuming no heat loss. $c_{water} = 4200$ J kg⁻¹ °C⁻¹, $L_{f,ice} = 3.34 \times 10^5$ J kg⁻¹.

1. Convert all masses to kg: $m_{ice}=0.05$ kg, $m_{hot\text{ }water}=0.2$ kg
2. Heat gained by ice = heat to melt ice + heat to warm melted ice to final temperature $T$:
3. $$Q_{gained} = m_{ice}L_f + m_{ice}c(T - 0)$$
4. Heat lost by hot water cooling from 40°C to $T$:
5. $$Q_{lost} = m_{hot\text{ }water}c(40 - T)$$
6. Equate heat lost and heat gained:
7. $$0.2 \times 4200 (40-T) = 0.05 \times 3.34 \times 10^5 + 0.05 \times 4200 T$$
8. Expand and solve for $T$:
9. $$33600 - 840T = 16700 + 210T \\ 16900 = 1050T \\ T \approx 16.1^\circ \text{C}$$

*Calculator:* allowed

## Experimental Measurement for Practical Exams

CIE practical exams frequently ask to describe experiments to measure specific latent heat. The standard method uses an electrical heater to supply a known amount of energy, then measures the mass of substance that changes phase.

For measuring the specific latent heat of vaporization of water, an immersed electrical heater supplies energy to boiling water. The energy supplied is $VIt$, where $V$ = voltage, $I$ = current, $t$ = time. If $m$ is the mass of steam condensed, then assuming no heat loss: $VIt = mL_v$, so $L_v = \frac{VIt}{m}$.

**Worked example:** A 50 W heater is run for 10 minutes to boil water. 13 g of water is converted to steam. Calculate $L_v$ from this data.

1. Calculate total energy supplied by the heater:
2. $$E = Pt = 50 \times (10 \times 60) = 30000 \text{ J}$$
3. Convert mass of steam to kg:
4. $$m = 13 \text{ g} = 0.013 \text{ kg}$$
5. Rearrange $E=mL$ to solve for $L_v$:
6. $$L_v = \frac{E}{m} = \frac{30000}{0.013} \approx 2.3 \times 10^6 \text{ J kg}^{-1}$$

> **note**
>
> In real experiments, heat is lost to the surroundings. This means less mass is vaporized for the same energy input, so the calculated value of $L_v$ will be higher than the true value.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using mass in grams instead of kilograms
  - Why it fails: Specific latent heat is defined per kilogram, so using grams will give an energy result 1000 times too small
  - Correct: Always convert mass to kilograms before substituting into $Q=mL$ by dividing grams by 1000
- **Wrong:** Adding a temperature change term for the phase change itself
  - Why it fails: Temperature is constant during phase change, so no temperature change occurs for the phase transition
  - Correct: Only use $Q=mL$ for the phase change, add $Q=mc\Delta T$ separately for any temperature change before or after the phase change
- **Wrong:** Mixing up $L_f$ and $L_v$ values for water
  - Why it fails: More intermolecular bonds are broken when turning liquid to gas than solid to liquid, so $L_v$ is much larger than $L_f$
  - Correct: Remember $L_f \approx 3.3 \times 10^5$ J kg⁻¹ and $L_v \approx 2.3 \times 10^6$ J kg⁻¹, a full order of magnitude difference
- **Wrong:** Claiming experimental heat loss leads to a lower calculated $L$
  - Why it fails: Heat lost means less mass changes phase for the energy input, so the calculated value is higher than the true value
  - Correct: For electrical heating experiments, unaccounted heat loss always gives a higher calculated value of specific latent heat

## Cheatsheet

| Quantity | Symbol | Formula/Rule | Value for water |
| --- | --- | --- | --- |
| Specific latent heat | $L$ | $Q = mL$, units J kg⁻¹ | - |
| Latent heat of fusion | $L_f$ | Solid ↔ liquid phase change | $3.34 \times 10^5$ J kg⁻¹ |
| Latent heat of vaporization | $L_v$ | Liquid ↔ gas phase change | $2.26 \times 10^6$ J kg⁻¹ |
| Calorimetry principle | - | Heat lost = Heat gained (insulated system) | - |
| Electrical experiment | - | Energy supplied = $VIt = mL$ | - |

## What's next

Specific latent heat is a foundational thermal physics concept that underpins advanced topics including entropy, heat engines, and thermal energy transfer in engineering systems. It is regularly combined with specific heat capacity in mixed calculation questions across Papers 1, 2 and 3 of CIE 9702. Mastering this sub-topic prepares you for more complex calorimetry problems and practical assessment questions. Next, you can explore related thermal physics concepts or deepen your practical skills for A-Level exams.

- [Specific Heat Capacity](https://www.owlsprep.com/study/cie-9702-u20-specific-heat-capacity/)
- [First Law of Thermodynamics](https://www.owlsprep.com/study/cie-9702-u20-first-law-of-thermodynamics/)

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