What decides the energy needed

Definition

Internal energy

The total energy stored by the particles of a system in their kinetic and potential energy stores.

When a material is heated and stays in the same state, the energy transferred increases its internal energy and raises its temperature.
The energy needed for this depends on three things: the mass of the material, its specific heat capacity and the change in temperature.
Doubling the mass doubles the energy needed, because there are twice as many particles to speed up.
Doubling the temperature rise doubles the energy needed, because each particle has to gain twice as much kinetic energy.
A material with double the specific heat capacity needs double the energy for the same mass and the same temperature rise.
A kettle filled to the top takes several times longer to boil than one holding just enough water for a single mug, because the energy needed is proportional to the mass of water.

The equation

Definition

Specific heat capacity

The energy needed to raise the temperature of $kg1\ \text{kg}$ of a substance by $∘C1\ ^\circ\text{C}$ .

The three factors combine into one equation: $temperature\text{change in thermal energy} = \text{mass} \times \text{specific heat capacity} \times \text{change in temperature}$ .
In symbols, this is $ΔE=m c Δθ\Delta E = m\,c\,\Delta\theta$ $Δ E = m c Δ θ$ .
1. $ΔE\Delta E$ is the change in thermal energy in joules, $J\text{J}$ .
2. $m$ is the mass in kilograms, $kg\text{kg}$ .
3. $c$ is the specific heat capacity in $∘C\text{J/kg}\,^\circ\text{C}$ .
4. $Δθ\Delta\theta$ is the change in temperature in $∘C^\circ\text{C}$ , where $θ\theta$ is the Greek letter theta.
The symbol $Δ\Delta$ means change in, so $Δθ\Delta\theta$ is always worked out as the final temperature minus the starting temperature.
You do not have to recall this equation from memory, but you must be able to choose it from a list of equations and apply it correctly.
The equation applies only when there is no change of state, because the temperature does not change while a substance is melting or boiling.

Calculating the energy change

Start by writing the equation, then work out $Δθ\Delta\theta$ if the question gives a starting and a final temperature.
Convert the mass into kilograms, remembering that $g1\ \text{kg} = 1000\ \text{g}$ .
Substitute the values, calculate, and finish with the unit, which for energy is the joule, $J\text{J}$ .
Large answers can be written in standard form or in kilojoules, where $J1\ \text{kJ} = 1000\ \text{J}$ .
When a material cools down, $Δθ\Delta\theta$ is negative and so $ΔE\Delta E$ comes out negative, which shows that the internal energy has decreased as energy is transferred to the surroundings.

Example

Heating water in a kettle

A kettle heats $kg1.5\ \text{kg}$ of water from $∘C15\ ^\circ\text{C}$ to $∘C100\ ^\circ\text{C}$ , and the specific heat capacity of water is $∘C4200\ \text{J/kg}\,^\circ\text{C}$ .
Write the equation.

\Delta E = m\,c\,\Delta\theta

Work out the temperature change.

\Delta\theta = 100 - 15 = 85\ ^\circ\text{C}

Substitute the values and calculate.

\Delta E = 1.5 \times 4200 \times 85 = 535\,500\ \text{J}

The internal energy of the water increases by $5.4 \times 10^{5}\ \text{J}$ , which is about $kJ540\ \text{kJ}$ .

Example

Mass in grams and a falling temperature

A $g250\ \text{g}$ aluminium baking tray cools from $∘C180\ ^\circ\text{C}$ to $∘C20\ ^\circ\text{C}$ , and the specific heat capacity of aluminium is $∘C900\ \text{J/kg}\,^\circ\text{C}$ .
Convert the mass into kilograms.

\dfrac{250}{1000} = 0.25\ \text{kg}

Work out the temperature change, which is negative because the tray is cooling.

\Delta\theta = 20 - 180 = -160\ ^\circ\text{C}

Substitute the values and calculate.

\Delta E = 0.25 \times 900 \times (-160) = -36\,000\ \text{J}

The negative sign shows that the internal energy of the tray falls by $J36\,000\ \text{J}$ as energy is transferred to the kitchen.

Rearranging the equation

Questions can give you the energy and ask for one of the other quantities instead.
To find the specific heat capacity, divide both sides by $Δθm\,\Delta\theta$ to give $\dfrac{\Delta E}{m\,\Delta\theta}$ .
To find the mass, divide both sides by $Δθc\,\Delta\theta$ to give $\dfrac{\Delta E}{c\,\Delta\theta}$ .
To find the temperature change, divide both sides by $cm\,c$ to give $c\Delta\theta = \dfrac{\Delta E}{m\,c}$ .
If the question asks for a final temperature, find $Δθ\Delta\theta$ first and then add it to the starting temperature, or subtract it if the material is cooling.
In a calculation with several steps, keep the unrounded value in your calculator and round only the final answer, usually to the same number of significant figures as the data.

Example

Finding a final temperature

An immersion heater transfers $J63\,000\ \text{J}$ to $kg0.50\ \text{kg}$ of water at $∘C18\ ^\circ\text{C}$ , and the specific heat capacity of water is $∘C4200\ \text{J/kg}\,^\circ\text{C}$ .
Rearrange the equation to make $Δθ\Delta\theta$ the subject.

\Delta\theta = \dfrac{\Delta E}{m\,c}

Substitute the values and calculate.

\Delta\theta = \dfrac{63\,000}{0.50 \times 4200} = 30\ ^\circ\text{C}

The water is being heated, so the temperature change is added to the starting temperature.

\text{final temperature} = 18 + 30 = 48\ ^\circ\text{C}

Example

Finding a specific heat capacity

A $kg0.80\ \text{kg}$ block of an unknown metal gains $J15\,600\ \text{J}$ , and its temperature rises from $∘C20\ ^\circ\text{C}$ to $∘C70\ ^\circ\text{C}$ .
Work out the temperature change.

\Delta\theta = 70 - 20 = 50\ ^\circ\text{C}

Rearrange the equation, substitute the values and calculate.

\dfrac{\Delta E}{m\,\Delta\theta} = \dfrac{15\,600}{0.80 \times 50} = 390\ \text{J/kg}\,^\circ\text{C}

This value matches the specific heat capacity of copper, so the block is probably made of copper.

Common Mistake

Do not substitute the final temperature in place of $Δθ\Delta\theta$ , because the equation needs the change in temperature.
Do not use a mass in grams, because the unit of $c$ is per kilogram and the answer would come out $1000$ times too big.
Do not use this equation for melting or boiling, because the temperature is not changing during a change of state.
Do not write the unit of specific heat capacity as $J/kg\text{J/kg}$ , because that is the unit of specific latent heat.

Exam technique

Showing your working

Calculations on this equation are usually worth two to four marks, and most of those marks are for the method rather than the final number.
Write the equation first, then show the substitution with every value in the correct unit, because a correct substitution earns credit even if the arithmetic then slips.
In a two-step question, a wrong value from the first step is carried forward, so finishing the method still scores the later marks.
Check that the size of your answer makes sense, since warming a mug of tea needs tens of thousands of joules rather than millions or just a handful.
Give the unit with the final answer, using the unit printed on the answer line where there is one.

Self review

Write the equation that links change in thermal energy, mass, specific heat capacity and change in temperature.
State the three factors that decide how much energy is needed to warm a material.
Calculate the energy needed to warm $kg2.0\ \text{kg}$ of water from $∘C20\ ^\circ\text{C}$ to $∘C50\ ^\circ\text{C}$ , using $4200\ \text{J/kg}\,^\circ\text{C}$ .
Rearrange $Δθ\Delta E = m\,c\,\Delta\theta$ to make $Δθ\Delta\theta$ the subject.
Explain why $ΔE\Delta E$ comes out negative when a material cools.

What decides the energy needed

Definition

Internal energy

The total energy stored by the particles of a system in their kinetic and potential energy stores.

When a material is heated and stays in the same state, the energy transferred increases its internal energy and raises its temperature.
The energy needed for this depends on three things: the mass of the material, its specific heat capacity and the change in temperature.
Doubling the mass doubles the energy needed, because there are twice as many particles to speed up.
Doubling the temperature rise doubles the energy needed, because each particle has to gain twice as much kinetic energy.
A material with double the specific heat capacity needs double the energy for the same mass and the same temperature rise.
A kettle filled to the top takes several times longer to boil than one holding just enough water for a single mug, because the energy needed is proportional to the mass of water.

The equation

Definition

Specific heat capacity

The energy needed to raise the temperature of $kg1\ \text{kg}$ of a substance by $∘C1\ ^\circ\text{C}$ .

The three factors combine into one equation: $temperature\text{change in thermal energy} = \text{mass} \times \text{specific heat capacity} \times \text{change in temperature}$ .
In symbols, this is $ΔE=m c Δθ\Delta E = m\,c\,\Delta\theta$ $Δ E = m c Δ θ$ .
1. $ΔE\Delta E$ is the change in thermal energy in joules, $J\text{J}$ .
2. $m$ is the mass in kilograms, $kg\text{kg}$ .
3. $c$ is the specific heat capacity in $∘C\text{J/kg}\,^\circ\text{C}$ .
4. $Δθ\Delta\theta$ is the change in temperature in $∘C^\circ\text{C}$ , where $θ\theta$ is the Greek letter theta.
The symbol $Δ\Delta$ means change in, so $Δθ\Delta\theta$ is always worked out as the final temperature minus the starting temperature.
You do not have to recall this equation from memory, but you must be able to choose it from a list of equations and apply it correctly.
The equation applies only when there is no change of state, because the temperature does not change while a substance is melting or boiling.

Calculating the energy change

Start by writing the equation, then work out $Δθ\Delta\theta$ if the question gives a starting and a final temperature.
Convert the mass into kilograms, remembering that $g1\ \text{kg} = 1000\ \text{g}$ .
Substitute the values, calculate, and finish with the unit, which for energy is the joule, $J\text{J}$ .
Large answers can be written in standard form or in kilojoules, where $J1\ \text{kJ} = 1000\ \text{J}$ .
When a material cools down, $Δθ\Delta\theta$ is negative and so $ΔE\Delta E$ comes out negative, which shows that the internal energy has decreased as energy is transferred to the surroundings.

Example

Heating water in a kettle

A kettle heats $kg1.5\ \text{kg}$ of water from $∘C15\ ^\circ\text{C}$ to $∘C100\ ^\circ\text{C}$ , and the specific heat capacity of water is $∘C4200\ \text{J/kg}\,^\circ\text{C}$ .
Write the equation.

\Delta E = m\,c\,\Delta\theta

Work out the temperature change.

\Delta\theta = 100 - 15 = 85\ ^\circ\text{C}

Substitute the values and calculate.

\Delta E = 1.5 \times 4200 \times 85 = 535\,500\ \text{J}

The internal energy of the water increases by $5.4 \times 10^{5}\ \text{J}$ , which is about $kJ540\ \text{kJ}$ .

Example

Mass in grams and a falling temperature

A $g250\ \text{g}$ aluminium baking tray cools from $∘C180\ ^\circ\text{C}$ to $∘C20\ ^\circ\text{C}$ , and the specific heat capacity of aluminium is $∘C900\ \text{J/kg}\,^\circ\text{C}$ .
Convert the mass into kilograms.

\dfrac{250}{1000} = 0.25\ \text{kg}

Work out the temperature change, which is negative because the tray is cooling.

\Delta\theta = 20 - 180 = -160\ ^\circ\text{C}

Substitute the values and calculate.

\Delta E = 0.25 \times 900 \times (-160) = -36\,000\ \text{J}

The negative sign shows that the internal energy of the tray falls by $J36\,000\ \text{J}$ as energy is transferred to the kitchen.

Rearranging the equation

Questions can give you the energy and ask for one of the other quantities instead.
To find the specific heat capacity, divide both sides by $Δθm\,\Delta\theta$ to give $\dfrac{\Delta E}{m\,\Delta\theta}$ .
To find the mass, divide both sides by $Δθc\,\Delta\theta$ to give $\dfrac{\Delta E}{c\,\Delta\theta}$ .
To find the temperature change, divide both sides by $cm\,c$ to give $c\Delta\theta = \dfrac{\Delta E}{m\,c}$ .
If the question asks for a final temperature, find $Δθ\Delta\theta$ first and then add it to the starting temperature, or subtract it if the material is cooling.
In a calculation with several steps, keep the unrounded value in your calculator and round only the final answer, usually to the same number of significant figures as the data.

Example

Finding a final temperature

An immersion heater transfers $J63\,000\ \text{J}$ to $kg0.50\ \text{kg}$ of water at $∘C18\ ^\circ\text{C}$ , and the specific heat capacity of water is $∘C4200\ \text{J/kg}\,^\circ\text{C}$ .
Rearrange the equation to make $Δθ\Delta\theta$ the subject.

\Delta\theta = \dfrac{\Delta E}{m\,c}

Substitute the values and calculate.

\Delta\theta = \dfrac{63\,000}{0.50 \times 4200} = 30\ ^\circ\text{C}

The water is being heated, so the temperature change is added to the starting temperature.

\text{final temperature} = 18 + 30 = 48\ ^\circ\text{C}

Example

Finding a specific heat capacity

A $kg0.80\ \text{kg}$ block of an unknown metal gains $J15\,600\ \text{J}$ , and its temperature rises from $∘C20\ ^\circ\text{C}$ to $∘C70\ ^\circ\text{C}$ .
Work out the temperature change.

\Delta\theta = 70 - 20 = 50\ ^\circ\text{C}

Rearrange the equation, substitute the values and calculate.

\dfrac{\Delta E}{m\,\Delta\theta} = \dfrac{15\,600}{0.80 \times 50} = 390\ \text{J/kg}\,^\circ\text{C}

This value matches the specific heat capacity of copper, so the block is probably made of copper.

Common Mistake

Do not substitute the final temperature in place of $Δθ\Delta\theta$ , because the equation needs the change in temperature.
Do not use a mass in grams, because the unit of $c$ is per kilogram and the answer would come out $1000$ times too big.
Do not use this equation for melting or boiling, because the temperature is not changing during a change of state.
Do not write the unit of specific heat capacity as $J/kg\text{J/kg}$ , because that is the unit of specific latent heat.

Exam technique

Showing your working

Calculations on this equation are usually worth two to four marks, and most of those marks are for the method rather than the final number.
Write the equation first, then show the substitution with every value in the correct unit, because a correct substitution earns credit even if the arithmetic then slips.
In a two-step question, a wrong value from the first step is carried forward, so finishing the method still scores the later marks.
Check that the size of your answer makes sense, since warming a mug of tea needs tens of thousands of joules rather than millions or just a handful.
Give the unit with the final answer, using the unit printed on the answer line where there is one.

Self review

Write the equation that links change in thermal energy, mass, specific heat capacity and change in temperature.
State the three factors that decide how much energy is needed to warm a material.
Calculate the energy needed to warm $kg2.0\ \text{kg}$ of water from $∘C20\ ^\circ\text{C}$ to $∘C50\ ^\circ\text{C}$ , using $4200\ \text{J/kg}\,^\circ\text{C}$ .
Rearrange $Δθ\Delta E = m\,c\,\Delta\theta$ to make $Δθ\Delta\theta$ the subject.
Explain why $ΔE\Delta E$ comes out negative when a material cools.

1.2.5 Internal energy and temperature change

What decides the energy needed

The equation

Calculating the energy change

Heating water in a kettle

Mass in grams and a falling temperature

Rearranging the equation

Finding a final temperature

Finding a specific heat capacity

Showing your working

Review 1.2.5 Internal energy and temperature change by teaching Genie

1.2.5 Internal energy and temperature change

What decides the energy needed

The equation

Calculating the energy change

Heating water in a kettle

Mass in grams and a falling temperature

Rearranging the equation

Finding a final temperature

Finding a specific heat capacity

Showing your working

Review 1.2.5 Internal energy and temperature change by teaching Genie

1.2.5 Internal energy and temperature change

What decides the energy needed

The equation

Calculating the energy change

Heating water in a kettle

Mass in grams and a falling temperature

Rearranging the equation

Finding a final temperature

Finding a specific heat capacity

Showing your working

Review 1.2.5 Internal energy and temperature change by teaching Genie

1.2.5 Internal energy and temperature change Revision Guide

1.2.5 Internal energy and temperature change

What decides the energy needed

The equation

Calculating the energy change

Heating water in a kettle

Mass in grams and a falling temperature

Rearranging the equation

Finding a final temperature

Finding a specific heat capacity

Showing your working

Review 1.2.5 Internal energy and temperature change by teaching Genie

1.2.5 Internal energy and temperature change Revision Guide