14.3 Specific Heat Capacity and Specific Latent Heat

1. Specific Heat Capacity

What is it?

When you heat something up, energy goes into it and its temperature rises. But different materials need different amounts of energy to reach the same temperature. For example, water needs much more energy to heat up than metal does.

Specific heat capacity (symbol: c) is the amount of energy needed to raise the temperature of 1 kg of a substance by 1 °C (or 1 K).

The SI unit of specific heat capacity is J kg⁻¹ K⁻¹ (joules per kilogram per kelvin).

The Formula

$Q = mc\Delta T$

Where:

• Q = energy transferred (in joules, J)
• m = mass of the substance (in kg)
• c = specific heat capacity (in J kg⁻¹ K⁻¹)
• ΔT = change in temperature (in °C or K — both give the same change)

You can rearrange this formula depending on what you need to find:

$c = \frac{Q}{m\Delta T} \qquad \Delta T = \frac{Q}{mc} \qquad m = \frac{Q}{c\Delta T}$

Worked Example

A block of aluminium of mass 950 g is heated from 24 °C to 80 °C. Calculate the energy required. (c for aluminium = 910 J kg⁻¹ K⁻¹)

Step 1: Convert mass to kg → 950 g = 0.950 kg

Step 2: Find the temperature change → ΔT = 80 − 24 = 56 °C

Step 3: Apply the formula:

Using a Heater

When an electric heater of power P runs for time t, the energy it supplies is: $Q = P \times t$

So the heating formula becomes: $P \times t = mc\Delta T$

This is very useful in experiments where a heater warms up a known mass of material.

Heat Exchange Between Two Objects

When a hot object is placed in contact with a cold one (and no heat escapes to the surroundings):

This allows you to set up an equation and solve for the final temperature.

Example: A hot metal ball (mass 200 g, temperature 200 °C, c = 800 J kg⁻¹ K⁻¹) is placed in water (mass 0.5 kg, temperature 25 °C, c = 4200 J kg⁻¹ K⁻¹). No heat is lost to surroundings. Find the final temperature T.

$(0.2)(800)(200 - T) = (0.5)(4200)(T - 25)$ $32000 - 160T = 2100T - 52500$ $84500 = 2260T$ $T \approx 37.4°C$