Temperature changes in a system and specific heat capacity

What you'll learn

How heating a system can increase its internal energy and temperature.
Why the temperature rise depends on energy input, mass, and the material.
How to use the equation $ΔE=mcΔθ\Delta E = mc\Delta\theta$ .
How to rearrange the equation and avoid the common unit mistakes.

The big idea: energy in, temperature up

When you heat an object, energy is transferred to it. If the object’s temperature increases, the particles in it have more energy in their internal energy store.

Definition

System and internal energy

A system is the object or group of objects you are focusing on.
Internal energy is the total energy stored by the particles in a system, including their kinetic energy and potential energy.

Temperature is a measure of how hot something is. In particle terms, a higher temperature usually means the particles have a greater average kinetic energy — they are moving or vibrating more energetically.

In this topic, we are focusing on situations where energy transfer causes a temperature change, not a change of state.

Common Mistake

Changes of state

The equation in this lesson is for temperature changes. During melting, boiling, freezing or condensing, energy can be transferred while the temperature stays constant, so a different idea is needed.

What affects the temperature change?

If the temperature of a system increases, the size of the temperature rise depends on three things:

the energy input to the system
the mass of the substance heated
the type of material, because different materials need different amounts of energy to warm up

The symbol $Δ\Delta$ means “change in”, so $Δθ\Delta\theta$ means change in temperature. The symbol $θ\theta$ is used for temperature in this equation.

Δθ=θfinal−θinitial\Delta\theta = \theta_{\text{final}} - \theta_{\text{initial}}

Key Idea

What controls the temperature rise?

For the same material, putting in more energy gives a bigger temperature rise. For the same energy input, a larger mass or a material with a larger specific heat capacity gives a smaller temperature rise.

The specific heat capacity equation

The relationship you need is:

ΔE=mcΔθ\Delta E = mc\Delta\theta

where:

$ΔE\Delta E$ is the change in thermal energy, in joules (J)
$m$ is the mass, in kilograms (kg)
$c$ is the specific heat capacity, in joules per kilogram per degree Celsius (J/kg °C)
$Δθ\Delta\theta$ is the temperature change, in degrees Celsius (°C)

Definition

Specific heat capacity

The specific heat capacity of a substance is the amount of energy required to raise the temperature of one kilogram of the substance by one degree Celsius.

A material with a high specific heat capacity needs a lot of energy for each kilogram to warm up by one degree Celsius. Water has a high specific heat capacity, which is why it takes quite a lot of energy to heat a full kettle.

Here is how the quantities in the equation link to a simple heating setup.

A labelled metal block specific heat capacity setup showing energy input, mass, temperature change and material

Example

Calculating energy transferred

A 0.50 kg aluminium block has a specific heat capacity of 900 J/kg °C. It is heated so that its temperature rises by 25 °C. Calculate the change in thermal energy.

Choose the equation because you know the mass, specific heat capacity and temperature change:
$ΔE=mcΔθ\Delta E = mc\Delta\theta$
Substitute the values with their units:
$∘C\Delta E = 0.50\,\text{kg} \times 900\,\text{J/kg }^\circ\text{C} \times 25\,^\circ\text{C}$
Calculate the energy change:
$J\Delta E = 11250\,\text{J}$
So the block gains 11,250 J of thermal energy.

Rearranging the equation

You will be given the equation on the Physics equation sheet, but you still need to know how to rearrange it.

Starting with:

ΔE=mcΔθ\Delta E = mc\Delta\theta

you can rearrange it to find different unknowns:

c=ΔEmΔθm=ΔEcΔθΔθ=ΔEmc\begin{aligned} c &= \frac{\Delta E}{m\Delta\theta} \\ m &= \frac{\Delta E}{c\Delta\theta} \\ \Delta\theta &= \frac{\Delta E}{mc} \end{aligned}

Tip

Before substituting

Convert mass into kilograms first. If the question gives grams, divide by 1000 before using the equation.

Example

Calculating specific heat capacity

A 0.20 kg metal block is supplied with 1500 J of energy. Its temperature rises from 18 °C to 28 °C. Calculate its specific heat capacity.

Find the temperature change:
$∘C\Delta\theta = 28\,^\circ\text{C} - 18\,^\circ\text{C} = 10\,^\circ\text{C}$
Rearrange the equation to make $c$ the subject:
$\frac{\Delta E}{m\Delta\theta}$
Substitute the values:
$\frac{1500\,\text{J}}{0.20\,\text{kg} \times 10\,^\circ\text{C}}$
Calculate:
$750\,\text{J/kg }^\circ\text{C}$
The specific heat capacity is 750 J/kg °C.

How mass and material change the result

The same equation can be written as:

Δθ=ΔEmc\Delta\theta = \frac{\Delta E}{mc}

This version is useful for understanding the pattern.

If $ΔE\Delta E$ increases, $Δθ\Delta\theta$ increases.
If $m$ increases, $Δθ\Delta\theta$ decreases.
If $c$ increases, $Δθ\Delta\theta$ decreases.

So a small metal block heats up more quickly than a large block of the same metal, if both receive the same energy. A material with a lower specific heat capacity also warms up more for the same energy input.

This graph shows the direct link between energy input and temperature change when the mass and material stay fixed.

A graph of temperature change against energy input showing steeper lines for smaller mass or lower specific heat capacity

Example

Comparing temperature rises

Two blocks are made from the same material and receive the same energy input. Block A has a mass of 0.50 kg. Block B has a mass of 1.00 kg. Compare their temperature rises.

Use the relationship:
$Δθ=ΔEmc\Delta\theta = \frac{\Delta E}{mc}$
Since both blocks have the same energy input and the same material, $ΔE\Delta E$ and $c$ are the same.
Compare the masses. Block B has twice the mass of Block A, so its denominator $m c$ is twice as large.
A larger denominator gives a smaller temperature rise. Block B’s temperature rise is half of Block A’s, so Block A warms up twice as much.

Common pitfalls

Common Mistake

Using the final temperature instead of the temperature change

In $ΔE=mcΔθ\Delta E = mc\Delta\theta$ , you must use the temperature change, not the final temperature. If something warms from 20 °C to 65 °C, then $Δθ\Delta\theta$ is 45 °C.

Common Mistake

Forgetting to convert grams to kilograms

The mass must be in kilograms because the unit of specific heat capacity is J/kg °C. For example, 250 g is 0.250 kg.

Tip

Sanity check

If you calculate a very large temperature rise for a large mass of water, pause and check your units. Water needs a lot of energy to warm up because its specific heat capacity is high.

Real heating is not perfectly efficient

In classroom experiments, not all the energy from a heater goes into the object you are trying to heat. Some energy is transferred to the surroundings, the thermometer, the heater itself, or the container.

In ideal GCSE calculations, unless the question says otherwise, assume the energy input causes the temperature change of the substance.

Exam technique

In the exam

Write down $ΔE=mcΔθ\Delta E = mc\Delta\theta$ and identify which quantity is unknown.
Convert mass into kg and calculate $Δθ\Delta\theta$ from final temperature minus initial temperature.
Substitute values with units, then check that your answer makes sense: bigger mass or bigger specific heat capacity should mean a smaller temperature rise for the same energy.

Self review

Check yourself

What does specific heat capacity tell you about a material?
Why does doubling the mass halve the temperature rise if the energy input and material stay the same?
A block warms from 22 °C to 57 °C. What value should you use for $Δθ\Delta\theta$ ?

Temperature changes in a system and specific heat capacity

What you'll learn

How heating a system can increase its internal energy and temperature.
Why the temperature rise depends on energy input, mass, and the material.
How to use the equation $ΔE=mcΔθ\Delta E = mc\Delta\theta$ .
How to rearrange the equation and avoid the common unit mistakes.

The big idea: energy in, temperature up

When you heat an object, energy is transferred to it. If the object’s temperature increases, the particles in it have more energy in their internal energy store.

Definition

System and internal energy

A system is the object or group of objects you are focusing on.
Internal energy is the total energy stored by the particles in a system, including their kinetic energy and potential energy.

In this topic, we are focusing on situations where energy transfer causes a temperature change, not a change of state.

Common Mistake

Changes of state

The equation in this lesson is for temperature changes. During melting, boiling, freezing or condensing, energy can be transferred while the temperature stays constant, so a different idea is needed.

What affects the temperature change?

If the temperature of a system increases, the size of the temperature rise depends on three things:

the energy input to the system
the mass of the substance heated
the type of material, because different materials need different amounts of energy to warm up

The symbol $Δ\Delta$ means “change in”, so $Δθ\Delta\theta$ means change in temperature. The symbol $θ\theta$ is used for temperature in this equation.

Δθ=θfinal−θinitial\Delta\theta = \theta_{\text{final}} - \theta_{\text{initial}}

Key Idea

What controls the temperature rise?

The specific heat capacity equation

The relationship you need is:

ΔE=mcΔθ\Delta E = mc\Delta\theta

where:

$ΔE\Delta E$ is the change in thermal energy, in joules (J)
$m$ is the mass, in kilograms (kg)
$c$ is the specific heat capacity, in joules per kilogram per degree Celsius (J/kg °C)
$Δθ\Delta\theta$ is the temperature change, in degrees Celsius (°C)

Definition

Specific heat capacity

The specific heat capacity of a substance is the amount of energy required to raise the temperature of one kilogram of the substance by one degree Celsius.

Here is how the quantities in the equation link to a simple heating setup.

A labelled metal block specific heat capacity setup showing energy input, mass, temperature change and material

Example

Calculating energy transferred

A 0.50 kg aluminium block has a specific heat capacity of 900 J/kg °C. It is heated so that its temperature rises by 25 °C. Calculate the change in thermal energy.

Choose the equation because you know the mass, specific heat capacity and temperature change:
$ΔE=mcΔθ\Delta E = mc\Delta\theta$
Substitute the values with their units:
$∘C\Delta E = 0.50\,\text{kg} \times 900\,\text{J/kg }^\circ\text{C} \times 25\,^\circ\text{C}$
Calculate the energy change:
$J\Delta E = 11250\,\text{J}$
So the block gains 11,250 J of thermal energy.

Rearranging the equation

You will be given the equation on the Physics equation sheet, but you still need to know how to rearrange it.

Starting with:

ΔE=mcΔθ\Delta E = mc\Delta\theta

you can rearrange it to find different unknowns:

c=ΔEmΔθm=ΔEcΔθΔθ=ΔEmc\begin{aligned} c &= \frac{\Delta E}{m\Delta\theta} \\ m &= \frac{\Delta E}{c\Delta\theta} \\ \Delta\theta &= \frac{\Delta E}{mc} \end{aligned}

Tip

Before substituting

Convert mass into kilograms first. If the question gives grams, divide by 1000 before using the equation.

Example

Calculating specific heat capacity

A 0.20 kg metal block is supplied with 1500 J of energy. Its temperature rises from 18 °C to 28 °C. Calculate its specific heat capacity.

Find the temperature change:
$∘C\Delta\theta = 28\,^\circ\text{C} - 18\,^\circ\text{C} = 10\,^\circ\text{C}$
Rearrange the equation to make $c$ the subject:
$\frac{\Delta E}{m\Delta\theta}$
Substitute the values:
$\frac{1500\,\text{J}}{0.20\,\text{kg} \times 10\,^\circ\text{C}}$
Calculate:
$750\,\text{J/kg }^\circ\text{C}$
The specific heat capacity is 750 J/kg °C.

How mass and material change the result

The same equation can be written as:

Δθ=ΔEmc\Delta\theta = \frac{\Delta E}{mc}

This version is useful for understanding the pattern.

If $ΔE\Delta E$ increases, $Δθ\Delta\theta$ increases.
If $m$ increases, $Δθ\Delta\theta$ decreases.
If $c$ increases, $Δθ\Delta\theta$ decreases.

This graph shows the direct link between energy input and temperature change when the mass and material stay fixed.

A graph of temperature change against energy input showing steeper lines for smaller mass or lower specific heat capacity

Example

Comparing temperature rises

Two blocks are made from the same material and receive the same energy input. Block A has a mass of 0.50 kg. Block B has a mass of 1.00 kg. Compare their temperature rises.

Use the relationship:
$Δθ=ΔEmc\Delta\theta = \frac{\Delta E}{mc}$
Since both blocks have the same energy input and the same material, $ΔE\Delta E$ and $c$ are the same.
Compare the masses. Block B has twice the mass of Block A, so its denominator $m c$ is twice as large.
A larger denominator gives a smaller temperature rise. Block B’s temperature rise is half of Block A’s, so Block A warms up twice as much.

Common pitfalls

Common Mistake

Using the final temperature instead of the temperature change

In $ΔE=mcΔθ\Delta E = mc\Delta\theta$ , you must use the temperature change, not the final temperature. If something warms from 20 °C to 65 °C, then $Δθ\Delta\theta$ is 45 °C.

Common Mistake

Forgetting to convert grams to kilograms

The mass must be in kilograms because the unit of specific heat capacity is J/kg °C. For example, 250 g is 0.250 kg.

Tip

Sanity check

If you calculate a very large temperature rise for a large mass of water, pause and check your units. Water needs a lot of energy to warm up because its specific heat capacity is high.

Real heating is not perfectly efficient

In ideal GCSE calculations, unless the question says otherwise, assume the energy input causes the temperature change of the substance.

Exam technique

In the exam

Write down $ΔE=mcΔθ\Delta E = mc\Delta\theta$ and identify which quantity is unknown.
Convert mass into kg and calculate $Δθ\Delta\theta$ from final temperature minus initial temperature.
Substitute values with units, then check that your answer makes sense: bigger mass or bigger specific heat capacity should mean a smaller temperature rise for the same energy.

Self review

Check yourself

What does specific heat capacity tell you about a material?
Why does doubling the mass halve the temperature rise if the energy input and material stay the same?
A block warms from 22 °C to 57 °C. What value should you use for $Δθ\Delta\theta$ ?

Temperature changes in a system and specific heat capacity

What you'll learn

The big idea: energy in, temperature up

What affects the temperature change?

The specific heat capacity equation

Rearranging the equation

How mass and material change the result

Common pitfalls

Real heating is not perfectly efficient

You've reached the end

Temperature changes in a system and specific heat capacity

What you'll learn

The big idea: energy in, temperature up

What affects the temperature change?

The specific heat capacity equation

Rearranging the equation

How mass and material change the result

Common pitfalls

Real heating is not perfectly efficient

You've reached the end

Temperature changes in a system and specific heat capacity

What you'll learn

The big idea: energy in, temperature up

What affects the temperature change?

The specific heat capacity equation

Rearranging the equation

How mass and material change the result

Common pitfalls

Real heating is not perfectly efficient

You've reached the end

Temperature changes in a system and specific heat capacity Revision Guide

Temperature changes in a system and specific heat capacity

What you'll learn

The big idea: energy in, temperature up

What affects the temperature change?

The specific heat capacity equation

Rearranging the equation

How mass and material change the result

Common pitfalls

Real heating is not perfectly efficient

You've reached the end

Temperature changes in a system and specific heat capacity Revision Guide