The thermal energy equation

Definition

Specific heat capacity

Specific heat capacity is the energy needed to raise the temperature of one kilogram of a substance by one degree Celsius.

The equation is $ΔQ=m×c×Δθ\Delta Q=m\times c\times\Delta\theta$ , which links the energy transferred to the mass, the material and the temperature change.
In this equation $ΔQ\Delta Q$ is the change in thermal energy in joules, $J\text{J}$ , $m$ is the mass in kilograms, $kg\text{kg}$ , $c$ is the specific heat capacity in $∘C\text{J/kg}\,^\circ\text{C}$ , and $Δθ\Delta\theta$ is the change in temperature in $∘C^\circ\text{C}$ .
The symbol $Δ\Delta$ means change in, and $θ\theta$ is the Greek letter theta, used here for temperature.
Work out the temperature change as $Δθ=θfinal−θstart\Delta\theta=\theta_{\text{final}}-\theta_{\text{start}}$ , and never substitute a single temperature reading on its own.
The equation only applies while the substance stays in the same state, because a change of state produces no temperature change to substitute.
Convert the mass into kilograms before substituting, so $g250\ \text{g}$ becomes $kg0.250\ \text{kg}$ .
Cooling uses the same equation, and the energy released when a substance cools by a given amount equals the energy needed to warm it by the same amount.
Energy transferred grows with all three factors, so a large mass, a high specific heat capacity or a big temperature change each increase the energy needed.

Rearranging and supplying the energy

Rearranging for the specific heat capacity gives $Δθc=\dfrac{\Delta Q}{m\,\Delta\theta}$ .
Rearranging for the mass gives $Δθm=\dfrac{\Delta Q}{c\,\Delta\theta}$ .
Rearranging for the temperature change gives $c\Delta\theta=\dfrac{\Delta Q}{m\,c}$ .
When an electric heater supplies the energy, the energy it transfers is $E=P×tE=P\times t$ , where $P$ is the power in watts and $t$ is the time in seconds.
If the potential difference and current are given instead, use $E=V×I×tE=V\times I\times t$ to find the energy supplied.
Setting the energy supplied equal to $ΔQ\Delta Q$ assumes that no energy escapes to the surroundings or into the container, which is worth stating as an assumption.
Because some energy always escapes in practice, an experimental value of $c$ usually comes out higher than the accepted value, since more energy was supplied than the substance actually gained.

Example

Energy needed to heat water

A kettle heats $kg0.50\ \text{kg}$ of water from $∘C20\,^\circ\text{C}$ to $∘C90\,^\circ\text{C}$ , and the specific heat capacity of water is $∘C4200\ \text{J/kg}\,^\circ\text{C}$ .
The temperature change is $∘C\Delta\theta=90-20=70\,^\circ\text{C}$ .
Using $Δθ\Delta Q=m\,c\,\Delta\theta$ gives $ΔQ=0.50×4200×70\Delta Q=0.50\times4200\times70$ .
The energy transferred is $J\Delta Q=1.47\times10^{5}\ \text{J}$ .

Example

Finding a specific heat capacity

A metal block of mass $kg0.80\ \text{kg}$ gains $J21\,600\ \text{J}$ and its temperature rises by $∘C30\,^\circ\text{C}$ .
Rearranging gives $Δθc=\dfrac{\Delta Q}{m\,\Delta\theta}$ .
Substituting gives $60024c=\dfrac{21\,600}{0.80\times30}=\dfrac{21\,600}{24}$ .
The specific heat capacity is $∘C900\ \text{J/kg}\,^\circ\text{C}$ , which matches aluminium.

Example

Temperature rise from a heater

A $W50\ \text{W}$ immersion heater runs for $s300\ \text{s}$ in $kg0.40\ \text{kg}$ of water.
The energy supplied is $JE=P\times t=50\times300=15\,000\ \text{J}$ .
Assuming all of that energy reaches the water, $0000.40×4200\Delta\theta=\dfrac{\Delta Q}{m\,c}=\dfrac{15\,000}{0.40\times4200}$ .
The temperature rise is $∘C\Delta\theta=8.9\,^\circ\text{C}$ to two significant figures.
The real rise would be smaller than this, because some energy is transferred to the beaker and to the surrounding air.

Practical

Investigating the properties of water

Aim: to determine the specific heat capacity of water, and to obtain a temperature against time graph for melting ice.
Apparatus, first part: insulated beaker or copper calorimeter with a lid, water, $V12\ \text{V}$ immersion heater, low voltage power supply, joulemeter or a voltmeter, ammeter and stopwatch, thermometer or temperature probe, digital balance, stirrer, lagging and a heatproof mat.
Apparatus, second part: crushed ice, boiling tube or small beaker, temperature probe or thermometer, larger beaker of warm water as a water bath, tripod, gauze and Bunsen burner or an electric heater, stopwatch and a stirrer.
Variables, first part: the energy supplied is the independent variable, the temperature of the water is the dependent variable, and the mass of water, the heater power, the insulation and the starting temperature are controlled.
Variables, second part: time is the independent variable, temperature is the dependent variable, and the mass of ice, the heating rate, the container and the position of the probe are controlled.
Method, specific heat capacity of water:
- Measure the mass of the empty container, then the mass of the container plus water, and subtract to find the mass of water in kilograms.
- Push the heater and the temperature probe through holes in the lid so that both are fully immersed but touching neither the base nor the sides.
- Wrap the sides of the container in lagging and fit the lid, so that as little energy as possible escapes.
- Record the starting temperature, then zero the joulemeter and switch the heater on.
- Stir gently and record the energy supplied and the temperature at regular intervals, for example every $s60\ \text{s}$ .
- Use $E=V×I×tE=V\times I\times t$ to calculate the energy supplied instead if no joulemeter is available, reading the potential difference and current while the heater runs.
- Switch off after a temperature rise of roughly $∘C20\,^\circ\text{C}$ , then calculate $Δθ\Delta\theta$ and use $Δθc=\dfrac{\Delta Q}{m\,\Delta\theta}$ .
Method, temperature against time graph for melting ice:
- Fill the boiling tube with crushed ice and place the probe centrally in the ice rather than against the glass.
- Record the starting temperature, then stand the tube in a beaker of warm water so the ice is warmed gently and evenly.
- Start the stopwatch and record the temperature at fixed intervals, for example every $s30\ \text{s}$ , stirring gently so the contents stay at one temperature.
- Keep going through the whole of the melting stage and on until the liquid water has clearly begun to warm up.
- Plot temperature on the vertical axis against time on the horizontal axis and draw a smooth curve through the points.
Results, first part: the accepted specific heat capacity of water is about $∘C4200\ \text{J/kg}\,^\circ\text{C}$ , and a school result is usually larger because some of the energy supplied warms the container and escapes to the room.
Results, second part: the graph rises while the ice warms, flattens into a horizontal section at about $∘C0\,^\circ\text{C}$ while the ice melts, and rises again once all the ice has become water.
Interpreting the flat section: energy is still entering the ice throughout it, but that energy changes the arrangement of the molecules instead of raising their average kinetic energy.
Maths: plot energy supplied against temperature for the first part and take the gradient, which equals $m×cm\times c$ , so dividing the gradient by the mass gives the specific heat capacity.
Watch out: energy lost to the surroundings, energy absorbed by the container and the heater itself, evaporation from an uncovered surface, the lag of a thermometer behind the true temperature, and poor stirring all distort the results.
Improvements: lag the container and fit a lid, stir continuously, use a joulemeter rather than separate meter readings, take readings frequently with a data logger, start below room temperature and finish an equal amount above it so gains and losses cancel, and repeat to find a mean.
Safety: wear eye protection, never switch on an immersion heater unless it is fully submerged, keep electrical connections and the power supply away from water, handle hot water and the hot heater with care, use a heatproof mat, and turn the Bunsen burner off when it is not in use.

Exam technique

Setting out the calculation

Write $Δθ\Delta Q=m\,c\,\Delta\theta$ before substituting, then rearrange on a separate line so the method is visible even if the arithmetic slips.
Show the subtraction that gives $Δθ\Delta\theta$ , because using a single temperature instead of a change is the commonest error in this calculation.
Convert grams to kilograms first, and say so in your working when the question gives the mass in grams.
Calculate the energy supplied with $tE=P\,t$ or $tE=V\,I\,t$ as its own step when the question describes a heater.
Say that the experimental value is too high and give the reason, since a bare statement about energy loss without a direction of error rarely earns the mark.
Label the flat part of a temperature against time graph as the change of state and explain it using particle arrangement rather than just naming it.

Common Mistake

Do not put a temperature reading into the equation where the temperature change belongs.
Do not use this equation across a change of state, because it cannot account for energy transferred at constant temperature.
Do not leave the mass in grams, because doing so makes the answer a thousand times too small.
Do not claim the practical value comes out too low, because energy escaping to the surroundings makes the calculated specific heat capacity too high.
Do not switch on an immersion heater in air, because it will overheat and can crack the glass around it.

Self review

State the thermal energy equation and give the unit of each quantity in it.
Calculate the energy needed to raise the temperature of $kg2.0\ \text{kg}$ of water by $∘C15\,^\circ\text{C}$ .
Rearrange the equation to give the specific heat capacity.
Describe how the mass of water is found in the specific heat capacity practical.
Explain why the experimental value of the specific heat capacity of water is usually too high.
Describe the shape of the temperature against time graph obtained for melting ice.

The thermal energy equation

Definition

Specific heat capacity

Specific heat capacity is the energy needed to raise the temperature of one kilogram of a substance by one degree Celsius.

The equation is $ΔQ=m×c×Δθ\Delta Q=m\times c\times\Delta\theta$ , which links the energy transferred to the mass, the material and the temperature change.
In this equation $ΔQ\Delta Q$ is the change in thermal energy in joules, $J\text{J}$ , $m$ is the mass in kilograms, $kg\text{kg}$ , $c$ is the specific heat capacity in $∘C\text{J/kg}\,^\circ\text{C}$ , and $Δθ\Delta\theta$ is the change in temperature in $∘C^\circ\text{C}$ .
The symbol $Δ\Delta$ means change in, and $θ\theta$ is the Greek letter theta, used here for temperature.
Work out the temperature change as $Δθ=θfinal−θstart\Delta\theta=\theta_{\text{final}}-\theta_{\text{start}}$ , and never substitute a single temperature reading on its own.
The equation only applies while the substance stays in the same state, because a change of state produces no temperature change to substitute.
Convert the mass into kilograms before substituting, so $g250\ \text{g}$ becomes $kg0.250\ \text{kg}$ .
Cooling uses the same equation, and the energy released when a substance cools by a given amount equals the energy needed to warm it by the same amount.
Energy transferred grows with all three factors, so a large mass, a high specific heat capacity or a big temperature change each increase the energy needed.

Rearranging and supplying the energy

Rearranging for the specific heat capacity gives $Δθc=\dfrac{\Delta Q}{m\,\Delta\theta}$ .
Rearranging for the mass gives $Δθm=\dfrac{\Delta Q}{c\,\Delta\theta}$ .
Rearranging for the temperature change gives $c\Delta\theta=\dfrac{\Delta Q}{m\,c}$ .
When an electric heater supplies the energy, the energy it transfers is $E=P×tE=P\times t$ , where $P$ is the power in watts and $t$ is the time in seconds.
If the potential difference and current are given instead, use $E=V×I×tE=V\times I\times t$ to find the energy supplied.
Setting the energy supplied equal to $ΔQ\Delta Q$ assumes that no energy escapes to the surroundings or into the container, which is worth stating as an assumption.
Because some energy always escapes in practice, an experimental value of $c$ usually comes out higher than the accepted value, since more energy was supplied than the substance actually gained.

Example

Energy needed to heat water

A kettle heats $kg0.50\ \text{kg}$ of water from $∘C20\,^\circ\text{C}$ to $∘C90\,^\circ\text{C}$ , and the specific heat capacity of water is $∘C4200\ \text{J/kg}\,^\circ\text{C}$ .
The temperature change is $∘C\Delta\theta=90-20=70\,^\circ\text{C}$ .
Using $Δθ\Delta Q=m\,c\,\Delta\theta$ gives $ΔQ=0.50×4200×70\Delta Q=0.50\times4200\times70$ .
The energy transferred is $J\Delta Q=1.47\times10^{5}\ \text{J}$ .

Example

Finding a specific heat capacity

A metal block of mass $kg0.80\ \text{kg}$ gains $J21\,600\ \text{J}$ and its temperature rises by $∘C30\,^\circ\text{C}$ .
Rearranging gives $Δθc=\dfrac{\Delta Q}{m\,\Delta\theta}$ .
Substituting gives $60024c=\dfrac{21\,600}{0.80\times30}=\dfrac{21\,600}{24}$ .
The specific heat capacity is $∘C900\ \text{J/kg}\,^\circ\text{C}$ , which matches aluminium.

Example

Temperature rise from a heater

A $W50\ \text{W}$ immersion heater runs for $s300\ \text{s}$ in $kg0.40\ \text{kg}$ of water.
The energy supplied is $JE=P\times t=50\times300=15\,000\ \text{J}$ .
Assuming all of that energy reaches the water, $0000.40×4200\Delta\theta=\dfrac{\Delta Q}{m\,c}=\dfrac{15\,000}{0.40\times4200}$ .
The temperature rise is $∘C\Delta\theta=8.9\,^\circ\text{C}$ to two significant figures.
The real rise would be smaller than this, because some energy is transferred to the beaker and to the surrounding air.

Practical

Investigating the properties of water

Aim: to determine the specific heat capacity of water, and to obtain a temperature against time graph for melting ice.
Apparatus, first part: insulated beaker or copper calorimeter with a lid, water, $V12\ \text{V}$ immersion heater, low voltage power supply, joulemeter or a voltmeter, ammeter and stopwatch, thermometer or temperature probe, digital balance, stirrer, lagging and a heatproof mat.
Apparatus, second part: crushed ice, boiling tube or small beaker, temperature probe or thermometer, larger beaker of warm water as a water bath, tripod, gauze and Bunsen burner or an electric heater, stopwatch and a stirrer.
Variables, first part: the energy supplied is the independent variable, the temperature of the water is the dependent variable, and the mass of water, the heater power, the insulation and the starting temperature are controlled.
Variables, second part: time is the independent variable, temperature is the dependent variable, and the mass of ice, the heating rate, the container and the position of the probe are controlled.
Method, specific heat capacity of water:
- Measure the mass of the empty container, then the mass of the container plus water, and subtract to find the mass of water in kilograms.
- Push the heater and the temperature probe through holes in the lid so that both are fully immersed but touching neither the base nor the sides.
- Wrap the sides of the container in lagging and fit the lid, so that as little energy as possible escapes.
- Record the starting temperature, then zero the joulemeter and switch the heater on.
- Stir gently and record the energy supplied and the temperature at regular intervals, for example every $s60\ \text{s}$ .
- Use $E=V×I×tE=V\times I\times t$ to calculate the energy supplied instead if no joulemeter is available, reading the potential difference and current while the heater runs.
- Switch off after a temperature rise of roughly $∘C20\,^\circ\text{C}$ , then calculate $Δθ\Delta\theta$ and use $Δθc=\dfrac{\Delta Q}{m\,\Delta\theta}$ .
Method, temperature against time graph for melting ice:
- Fill the boiling tube with crushed ice and place the probe centrally in the ice rather than against the glass.
- Record the starting temperature, then stand the tube in a beaker of warm water so the ice is warmed gently and evenly.
- Start the stopwatch and record the temperature at fixed intervals, for example every $s30\ \text{s}$ , stirring gently so the contents stay at one temperature.
- Keep going through the whole of the melting stage and on until the liquid water has clearly begun to warm up.
- Plot temperature on the vertical axis against time on the horizontal axis and draw a smooth curve through the points.
Results, first part: the accepted specific heat capacity of water is about $∘C4200\ \text{J/kg}\,^\circ\text{C}$ , and a school result is usually larger because some of the energy supplied warms the container and escapes to the room.
Results, second part: the graph rises while the ice warms, flattens into a horizontal section at about $∘C0\,^\circ\text{C}$ while the ice melts, and rises again once all the ice has become water.
Interpreting the flat section: energy is still entering the ice throughout it, but that energy changes the arrangement of the molecules instead of raising their average kinetic energy.
Maths: plot energy supplied against temperature for the first part and take the gradient, which equals $m×cm\times c$ , so dividing the gradient by the mass gives the specific heat capacity.
Watch out: energy lost to the surroundings, energy absorbed by the container and the heater itself, evaporation from an uncovered surface, the lag of a thermometer behind the true temperature, and poor stirring all distort the results.
Improvements: lag the container and fit a lid, stir continuously, use a joulemeter rather than separate meter readings, take readings frequently with a data logger, start below room temperature and finish an equal amount above it so gains and losses cancel, and repeat to find a mean.
Safety: wear eye protection, never switch on an immersion heater unless it is fully submerged, keep electrical connections and the power supply away from water, handle hot water and the hot heater with care, use a heatproof mat, and turn the Bunsen burner off when it is not in use.

Exam technique

Setting out the calculation

Write $Δθ\Delta Q=m\,c\,\Delta\theta$ before substituting, then rearrange on a separate line so the method is visible even if the arithmetic slips.
Show the subtraction that gives $Δθ\Delta\theta$ , because using a single temperature instead of a change is the commonest error in this calculation.
Convert grams to kilograms first, and say so in your working when the question gives the mass in grams.
Calculate the energy supplied with $tE=P\,t$ or $tE=V\,I\,t$ as its own step when the question describes a heater.
Say that the experimental value is too high and give the reason, since a bare statement about energy loss without a direction of error rarely earns the mark.
Label the flat part of a temperature against time graph as the change of state and explain it using particle arrangement rather than just naming it.

Common Mistake

Do not put a temperature reading into the equation where the temperature change belongs.
Do not use this equation across a change of state, because it cannot account for energy transferred at constant temperature.
Do not leave the mass in grams, because doing so makes the answer a thousand times too small.
Do not claim the practical value comes out too low, because energy escaping to the surroundings makes the calculated specific heat capacity too high.
Do not switch on an immersion heater in air, because it will overheat and can crack the glass around it.

Self review

State the thermal energy equation and give the unit of each quantity in it.
Calculate the energy needed to raise the temperature of $kg2.0\ \text{kg}$ of water by $∘C15\,^\circ\text{C}$ .
Rearrange the equation to give the specific heat capacity.
Describe how the mass of water is found in the specific heat capacity practical.
Explain why the experimental value of the specific heat capacity of water is usually too high.
Describe the shape of the temperature against time graph obtained for melting ice.

14.2.3 Specific heat capacity calculation

The thermal energy equation

Rearranging and supplying the energy

Energy needed to heat water

Finding a specific heat capacity

Temperature rise from a heater

Investigating the properties of water

Setting out the calculation

Review 14.2.3 Specific heat capacity calculation by teaching Genie

14.2.3 Specific heat capacity calculation

The thermal energy equation

Rearranging and supplying the energy

Energy needed to heat water

Finding a specific heat capacity

Temperature rise from a heater

Investigating the properties of water

Setting out the calculation

Review 14.2.3 Specific heat capacity calculation by teaching Genie

14.2.3 Specific heat capacity calculation

The thermal energy equation

Rearranging and supplying the energy

Energy needed to heat water

Finding a specific heat capacity

Temperature rise from a heater

Investigating the properties of water

Setting out the calculation

Review 14.2.3 Specific heat capacity calculation by teaching Genie

14.2.3 Specific heat capacity calculation Revision Guide

14.2.3 Specific heat capacity calculation

The thermal energy equation

Rearranging and supplying the energy

Energy needed to heat water

Finding a specific heat capacity

Temperature rise from a heater

Investigating the properties of water

Setting out the calculation

Review 14.2.3 Specific heat capacity calculation by teaching Genie

14.2.3 Specific heat capacity calculation Revision Guide