Heat capacity and specific heat capacity: WASSCE Physics

Put an empty iron pot and a pot of water on two identical fires. The empty pot is too hot to touch in a minute; the water is barely warm. Different substances need very different amounts of heat to warm up — and that difference is measured by specific heat capacity.

Learning objectives

By the end of this lesson you should be able to:

define heat capacity and specific heat capacity and state their units
use $Q = mc\Delta\theta$ and $Q = C\Delta\theta$
explain the importance of the high specific heat capacity of water
solve problems involving heaters, power and temperature rise

Heat capacity

Definition

The heat capacity $C$ of a body is the quantity of heat needed to raise the temperature of the whole body by 1 K (or 1 °C):

C = \dfrac{Q}{\Delta\theta} \quad (\text{unit: J K}^{-1})

\begin{gathered}C = \dfrac{Q}{\Delta\theta} \\[6pt] (\text{unit: J K}^{-1})\end{gathered}

Heat capacity belongs to a particular object: a large pot of water has a larger heat capacity than a cup of water.

Specific heat capacity

Definition

The specific heat capacity $c$ of a substance is the quantity of heat needed to raise the temperature of 1 kg of the substance by 1 K:

c = \dfrac{Q}{m\,\Delta\theta} \quad (\text{unit: J kg}^{-1}\ \text{K}^{-1})

\begin{gathered}c = \dfrac{Q}{m\,\Delta\theta} \\[6pt] (\text{unit: J kg}^{-1}\ \text{K}^{-1})\end{gathered}

So the heat absorbed (or given out) when a mass $m$ changes temperature by $\Delta\theta$ is

Q = mc\Delta\theta \qquad\text{and}\qquad C = mc

Substance	$c$ / J kg⁻¹ K⁻¹
Water	4200
Ice	2100
Aluminium	900
Glass	670
Iron / steel	450
Copper	400
Lead	130
Mercury	140

Worked example

How much heat is needed to raise the temperature of 2 kg of water from 25 °C to 100 °C? ( $c = 4200$ J kg⁻¹ K⁻¹)

Q = mc\Delta\theta = 2 \times 4200 \times 75 = 630\,000\ \text{J} = 630\ \text{kJ}

\begin{aligned}Q &= mc\Delta\theta \\ &= 2 \times 4200 \times 75 \\ &= 630\,000\ \text{J} \\ &= 630\ \text{kJ}\end{aligned}

Worked example

A 500 g copper block absorbs 4000 J of heat. By how much does its temperature rise? ( $c = 400$ J kg⁻¹ K⁻¹)

\Delta\theta = \dfrac{Q}{mc} = \dfrac{4000}{0.5 \times 400} = 20\ \text{K}

\begin{aligned}\Delta\theta &= \dfrac{Q}{mc} \\ &= \dfrac{4000}{0.5 \times 400} \\ &= 20\ \text{K}\end{aligned}

The same 4000 J would warm 0.5 kg of water by only $\dfrac{4000}{0.5 \times 4200} = 1.9$ K.

Common mistake

Always convert grams to kilograms before using $Q = mc\Delta\theta$ with $c$ in J kg⁻¹ K⁻¹. And $\Delta\theta$ is the change in temperature: a rise from 20 °C to 60 °C is 40 K, not 60.

Heating with an electric heater

An electric heater of power $P$ supplies energy $Pt$ in time $t$ . If all of it goes into the substance:

Pt = mc\Delta\theta

Worked example

A 2 kW electric kettle holds 1.5 kg of water at 20 °C. How long does it take to bring the water to 100 °C, assuming no heat loss?

t = \dfrac{mc\Delta\theta}{P} = \dfrac{1.5 \times 4200 \times 80}{2000} = 252\ \text{s} \approx 4.2\ \text{minutes}

\begin{aligned}t &= \dfrac{mc\Delta\theta}{P} \\ &= \dfrac{1.5 \times 4200 \times 80}{2000} \\ &= 252\ \text{s} \\ &\approx 4.2\ \text{minutes}\end{aligned}

In practice it takes a little longer, because some heat warms the kettle and escapes to the surroundings.

Why water's high specific heat capacity matters

Water has one of the highest specific heat capacities of any common substance. This means it absorbs a lot of heat for a small temperature rise and releases a lot as it cools.

Cooling systems: car radiators and generator engines use water to carry heat away.
Hot-water bottles and heating systems store a lot of heat in water.
Climate: the sea warms and cools slowly, so coastal places such as Freetown have milder temperature changes between day and night than inland places far from the sea.
Sea and land breezes are driven partly by land heating faster than sea (covered in heat transfer).
The body: being mostly water, it resists sudden temperature changes.

In Sierra Leone

At midday on Lumley Beach, the sand burns your feet while the sea is pleasantly cool. Sand has a much lower specific heat capacity than water, so the same sunshine raises its temperature far more.

Key points

Heat capacity $C = Q/\Delta\theta$ (J K⁻¹) belongs to an object; specific heat capacity $c = Q/(m\Delta\theta)$ (J kg⁻¹ K⁻¹) belongs to a substance.
$Q = mc\Delta\theta$ ; $C = mc$ .
With an electric heater (no losses): $Pt = mc\Delta\theta$ .
Water's high $c$ (4200 J kg⁻¹ K⁻¹) makes it an excellent coolant and moderates coastal climates.

WASSCE examination focus

WAEC tests this

Define heat capacity and specific heat capacity (the difference between them is often asked) and use $Q = mc\Delta\theta$ in calculations, including electrical heating ( $Pt = mc\Delta\theta$ ). Explain why water is used in car radiators. Watch units: grams to kilograms, kW to W, minutes to seconds.

Quick check

State the unit of specific heat capacity.
Find the heat needed to warm 0.5 kg of aluminium by 40 K ( $c = 900$ J kg⁻¹ K⁻¹).
Why is water used as a coolant in engines?

Answers: 1. J kg⁻¹ K⁻¹. 2. $0.5 \times 900 \times 40 = 18\,000$ J. 3. Its high specific heat capacity lets it absorb a lot of heat with only a small rise in temperature.

Heat capacity and specific heat capacity

Learning objectives

Heat capacity

Specific heat capacity

Heating with an electric heater

Why water's high specific heat capacity matters

Key points

WASSCE examination focus

Quick check

Keep going with quantity of heat and calorimetry

More lessons in Quantity of Heat and Calorimetry