Internal Energy and Specific Heat Capacity

Why does it take so long to boil a full kettle but only moments to warm a single spoonful of water? Why does the sand on a beach become scorching in the midday sun while the sea stays cool? The answer lies in specific heat capacity — a property that measures how much energy a material needs to change its temperature. Some materials warm up readily with only a little energy; others, like water, need a great deal. This lesson, part of Topic P1 (Matter) of OCR Gateway Science A, explains how heating raises a substance's internal energy, defines specific heat capacity, works through the equation $E = mc\Delta\theta$ in every direction, and walks through the required practical for measuring it.

By the end of this lesson you should be able to explain how heating raises internal energy, define specific heat capacity, use and rearrange $E = mc\Delta\theta$ , describe the required practical to determine a specific heat capacity, and identify the main sources of error in it.

Heating and Internal Energy

Recall that the internal energy of a substance is the total kinetic and potential energy of its particles. When you transfer energy to a substance by heating it, you raise its internal energy, and one of two things happens:

the particles move (or vibrate) faster, so the temperature rises (the energy has gone into the particles' kinetic energy); or
if the substance is at a melting or boiling point, the energy breaks the forces between particles and the substance changes state at constant temperature (the energy has gone into the particles' potential energy).

This lesson deals with the first case — heating that raises the temperature. How much the temperature rises for a given amount of energy depends on three things: the mass of the substance, the energy supplied, and a property of the material called its specific heat capacity.

Exam Tip: Be clear which "energy" is which. Heating raises internal energy; when this increases the particles' kinetic energy the temperature goes up. Temperature is a measure of the average kinetic energy of the particles, not the total internal energy.

What Is Specific Heat Capacity?

The specific heat capacity of a material is the energy needed to raise the temperature of $1 \text{ kg}$ of the material by $1\,°\text{C}$ (equivalently, by $1 \text{ K}$ — a change of one kelvin is the same size as a change of one degree Celsius). It is given the symbol $c$ and its unit is $\text{J/kg}\,°\text{C}$ (joules per kilogram per degree Celsius).

A material with a high specific heat capacity needs a lot of energy to warm up (and releases a lot when it cools); a material with a low specific heat capacity warms up and cools down easily. Water has an unusually high specific heat capacity of about $4200 \text{ J/kg}\,°\text{C}$ , which is why it is used in central-heating systems and car radiators (it can carry a lot of energy) and why coastal climates are milder (the sea stores energy in summer and releases it slowly in winter).

Material	Specific heat capacity / $\text{J/kg}\,°\text{C}$
Water	4200
Aluminium	900
Iron / steel	450
Copper	385
Lead	130

Exam Tip: Specific heat capacity is the energy to raise $1 \text{ kg}$ by $1\,°\text{C}$ . Water's high value (≈ $4200 \text{ J/kg}\,°\text{C}$ ) means it takes a lot of energy to heat and stores a lot of energy — link this to its use as a coolant and to mild coastal climates.

The Specific Heat Capacity Equation

The energy needed to change the temperature of a substance is given by:

$E = mc\Delta\theta$

where $E$ is the energy transferred (in $\text{J}$ ), $m$ is the mass (in $\text{kg}$ ), $c$ is the specific heat capacity (in $\text{J/kg}\,°\text{C}$ ), and $\Delta\theta$ is the change in temperature (in $°\text{C}$ ). The symbol $\Delta\theta$ (delta theta) means "the change in temperature" — always the final temperature minus the starting temperature.

The equation rearranges to make any quantity the subject:

$E = mc\Delta\theta \qquad c = \dfrac{E}{m\Delta\theta} \qquad \Delta\theta = \dfrac{E}{mc} \qquad m = \dfrac{E}{c\Delta\theta}$

Worked Example 1 — Energy to heat water

How much energy is needed to raise the temperature of $2 \text{ kg}$ of water from $20\,°\text{C}$ to $70\,°\text{C}$ ? (Specific heat capacity of water $= 4200 \text{ J/kg}\,°\text{C}$ .)

Step 1 — find the temperature change: $\Delta\theta = 70 - 20 = 50\,°\text{C}$ .

Step 2 — write the equation: $E = mc\Delta\theta$ .

Step 3 — substitute: $E = 2 \times 4200 \times 50$ .

Step 4 — calculate: $E = 420\,000 \text{ J}$ (which is $420 \text{ kJ}$ ).

Answer: $420\,000 \text{ J}$ .

Worked Example 2 — Energy to heat a metal

Calculate the energy needed to raise the temperature of a $0.5 \text{ kg}$ aluminium block by $30\,°\text{C}$ . (Specific heat capacity of aluminium $= 900 \text{ J/kg}\,°\text{C}$ .)

Step 1 — the temperature change is already given: $\Delta\theta = 30\,°\text{C}$ .

Step 2 — write the equation: $E = mc\Delta\theta$ .

Step 3 — substitute: $E = 0.5 \times 900 \times 30$ .

Step 4 — calculate: $E = 13\,500 \text{ J}$ .

Answer: $13\,500 \text{ J}$ (= $13.5 \text{ kJ}$ ).

Worked Example 3 — Rearranging for specific heat capacity

A $2 \text{ kg}$ block of metal is given $39\,000 \text{ J}$ of energy and its temperature rises by $50\,°\text{C}$ . Calculate the specific heat capacity of the metal and suggest what it might be.

Step 1 — rearrange for $c$ : $c = \dfrac{E}{m\Delta\theta}$ .

Step 2 — substitute: $c = \dfrac{39\,000}{2 \times 50} = \dfrac{39\,000}{100}$ .

Step 3 — calculate: $c = 390 \text{ J/kg}\,°\text{C}$ .

Answer: $c = 390 \text{ J/kg}\,°\text{C}$ , which is very close to the value for copper ( $385 \text{ J/kg}\,°\text{C}$ ).

Worked Example 4 — Rearranging for temperature change

A $1.5 \text{ kg}$ iron block is given $33\,750 \text{ J}$ of energy. By how much does its temperature rise? (Specific heat capacity of iron $= 450 \text{ J/kg}\,°\text{C}$ .)

Step 1 — rearrange for $\Delta\theta$ : $\Delta\theta = \dfrac{E}{mc}$ .

Step 2 — substitute: $\Delta\theta = \dfrac{33\,750}{1.5 \times 450} = \dfrac{33\,750}{675}$ .

Step 3 — calculate: $\Delta\theta = 50\,°\text{C}$ .

Answer: the temperature rises by $50\,°\text{C}$ .

Exam Tip: Always work out $\Delta\theta$ as final − start first, and make sure the mass is in kilograms (so $500 \text{ g}$ becomes $0.5 \text{ kg}$ ). Mixing up grams and kilograms is the commonest error in these calculations and gives an answer 1000 times too big or too small.

Required Practical: Determining a Specific Heat Capacity

A key P1 required practical is to measure the specific heat capacity of a material — usually a metal block, but the same method works for water. The idea is to supply a measured amount of electrical energy to a known mass and measure the resulting temperature rise, then calculate $c$ .

The apparatus is a metal block with two holes drilled in it: one for an electric heater (immersion heater) and one for a thermometer. The heater is connected through a joulemeter (or, equivalently, an ammeter and voltmeter together with a timer so the energy can be calculated).

Internal Energy and Specific Heat Capacity

Internal Energy and Specific Heat Capacity

Heating and Internal Energy

What Is Specific Heat Capacity?

The Specific Heat Capacity Equation

Worked Example 1 — Energy to heat water

Worked Example 2 — Energy to heat a metal

Worked Example 3 — Rearranging for specific heat capacity

Worked Example 4 — Rearranging for temperature change

Required Practical: Determining a Specific Heat Capacity

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