The latent heat equation

Definition

Specific latent heat

Specific latent heat is the energy needed to change the state of one kilogram of a substance without changing its temperature.

The equation is $Q=m×LQ=m\times L$ , which gives the energy involved in changing the state of a mass of substance.
In this equation $Q$ is the thermal energy for the change of state in joules, $J\text{J}$ , $m$ is the mass in kilograms, $kg\text{kg}$ , and $L$ is the specific latent heat in joules per kilogram, $J/kg\text{J/kg}$ .
There is no temperature term in the equation, because the temperature does not change while the state changes.
Use the specific latent heat of fusion for melting and for freezing, and the specific latent heat of vaporisation for boiling and for condensing.
The mass in the equation is only the mass that actually changes state, so half a kilogram melting out of a two kilogram block gives $kgm=0.5\ \text{kg}$ .
The same equation gives the energy released when a substance freezes or condenses, because the transfer runs the same way in reverse.
Values of $L$ are large, so answers are usually written in standard form, for example $J8.4\times10^{4}\ \text{J}$ rather than $J84\,000\ \text{J}$ .

Rearranging and supplying the energy

Rearranging for the mass gives $m=QLm=\dfrac{Q}{L}$ , which tells you how much of a substance a given amount of energy can melt or boil.
Rearranging for the specific latent heat gives $L=QmL=\dfrac{Q}{m}$ , which is the form used to find $L$ from experimental readings.
When an electric heater supplies the energy, calculate it from $E=P×tE=P\times t$ or from $E=V×I×tE=V\times I\times t$ and use that value as $Q$ .
Convert the mass to kilograms before substituting, because a mass in grams gives an answer that is out by a factor of a thousand.
Setting the energy supplied equal to $Q$ assumes that all of it reaches the substance, so an experimental value of $L$ is usually too high.

Problems with more than one stage

Many questions ask for the total energy to take a substance through a temperature change and a change of state, so split the problem into stages.
Use $Δθ\Delta Q=m\,c\,\Delta\theta$ for a stage in which the temperature changes and the state does not.
Use $LQ=m\,L$ for a stage in which the state changes and the temperature does not.
Add the energies for the separate stages together to get the total, and keep the working for each stage clearly labelled.
Check which state the substance is in for each stage, because the specific heat capacity of ice differs from that of liquid water.

Example

Energy needed to melt ice

A mass of $kg0.25\ \text{kg}$ of ice at $∘C0\,^\circ\text{C}$ is melted into water at $∘C0\,^\circ\text{C}$ , and the specific latent heat of fusion of water is $J/kg3.34\times10^{5}\ \text{J/kg}$ .
Using $LQ=m\,L$ gives $Q=0.25×3.34×105Q=0.25\times3.34\times10^{5}$ .
The energy needed is $JQ=8.35\times10^{4}\ \text{J}$ .
The temperature of the water is still $∘C0\,^\circ\text{C}$ after this transfer, because all of the energy went into the change of state.

Example

Finding a specific latent heat

A $W60\ \text{W}$ heater is switched on for $s500\ \text{s}$ in a funnel of ice at $∘C0\,^\circ\text{C}$ , and $kg0.090\ \text{kg}$ of water is collected.
The energy supplied is $JE=P\times t=60\times500=30\,000\ \text{J}$ .
Rearranging gives $0000.090L=\dfrac{Q}{m}=\dfrac{30\,000}{0.090}$ .
The specific latent heat of fusion works out as $J/kgL=3.3\times10^{5}\ \text{J/kg}$ to two significant figures.
Some ice would melt from the warmth of the room as well, which is why a second funnel of ice with no heater is used as a control.

Example

Heating water and then boiling it

A mass of $kg0.10\ \text{kg}$ of water at $∘C80\,^\circ\text{C}$ is heated to $∘C100\,^\circ\text{C}$ and then boiled away completely.
The first stage is a temperature change, so $J\Delta Q=m\,c\,\Delta\theta=0.10\times4200\times20=8400\ \text{J}$ .
The second stage is a change of state, so $JQ=m\,L=0.10\times2.26\times10^{6}=2.26\times10^{5}\ \text{J}$ .
The total energy is $J8400+2.26\times10^{5}=2.34\times10^{5}\ \text{J}$ .
Boiling the water accounts for about $96%96\%$ of the total, even though it produces no change in temperature.

Exam technique

Setting out the calculation

Write $LQ=m\,L$ before substituting, then rearrange on its own line when the question asks for the mass or for $L$ .
State which specific latent heat you are using, because a question involving both melting and boiling gives two values to choose between.
Label the two stages of a combined problem as separate calculations, so partial credit survives an arithmetic slip in one of them.
Give the answer in standard form with a unit, since latent heat answers run into hundreds of thousands of joules.
Name the energy escaping to the surroundings and say that it makes the value too high when a question asks you to comment on an experimental result.

Common Mistake

Do not include a temperature change in this equation, because $LQ=m\,L$ has no temperature term.
Do not use the specific latent heat of fusion for boiling, because the vaporisation value is far larger.
Do not use the total mass when only part of the substance changes state.
Do not answer a two stage question with a single calculation, because the marks are split between the stages.
Do not forget to convert the mass into kilograms, because the specific latent heat is quoted per kilogram.

Self review

State the latent heat equation and give the unit of each quantity in it.
Calculate the energy released when $kg0.40\ \text{kg}$ of water at $∘C0\,^\circ\text{C}$ freezes.
Rearrange the equation to find the mass melted by a known amount of energy.
Explain why the equation contains no temperature term.
Describe how you would find the total energy needed to melt ice and then warm the water produced.
Explain why an experimental value of specific latent heat usually comes out too high.

The latent heat equation

Definition

Specific latent heat

Specific latent heat is the energy needed to change the state of one kilogram of a substance without changing its temperature.

The equation is $Q=m×LQ=m\times L$ , which gives the energy involved in changing the state of a mass of substance.
In this equation $Q$ is the thermal energy for the change of state in joules, $J\text{J}$ , $m$ is the mass in kilograms, $kg\text{kg}$ , and $L$ is the specific latent heat in joules per kilogram, $J/kg\text{J/kg}$ .
There is no temperature term in the equation, because the temperature does not change while the state changes.
Use the specific latent heat of fusion for melting and for freezing, and the specific latent heat of vaporisation for boiling and for condensing.
The mass in the equation is only the mass that actually changes state, so half a kilogram melting out of a two kilogram block gives $kgm=0.5\ \text{kg}$ .
The same equation gives the energy released when a substance freezes or condenses, because the transfer runs the same way in reverse.
Values of $L$ are large, so answers are usually written in standard form, for example $J8.4\times10^{4}\ \text{J}$ rather than $J84\,000\ \text{J}$ .

Rearranging and supplying the energy

Rearranging for the mass gives $m=QLm=\dfrac{Q}{L}$ , which tells you how much of a substance a given amount of energy can melt or boil.
Rearranging for the specific latent heat gives $L=QmL=\dfrac{Q}{m}$ , which is the form used to find $L$ from experimental readings.
When an electric heater supplies the energy, calculate it from $E=P×tE=P\times t$ or from $E=V×I×tE=V\times I\times t$ and use that value as $Q$ .
Convert the mass to kilograms before substituting, because a mass in grams gives an answer that is out by a factor of a thousand.
Setting the energy supplied equal to $Q$ assumes that all of it reaches the substance, so an experimental value of $L$ is usually too high.

Problems with more than one stage

Many questions ask for the total energy to take a substance through a temperature change and a change of state, so split the problem into stages.
Use $Δθ\Delta Q=m\,c\,\Delta\theta$ for a stage in which the temperature changes and the state does not.
Use $LQ=m\,L$ for a stage in which the state changes and the temperature does not.
Add the energies for the separate stages together to get the total, and keep the working for each stage clearly labelled.
Check which state the substance is in for each stage, because the specific heat capacity of ice differs from that of liquid water.

Example

Energy needed to melt ice

A mass of $kg0.25\ \text{kg}$ of ice at $∘C0\,^\circ\text{C}$ is melted into water at $∘C0\,^\circ\text{C}$ , and the specific latent heat of fusion of water is $J/kg3.34\times10^{5}\ \text{J/kg}$ .
Using $LQ=m\,L$ gives $Q=0.25×3.34×105Q=0.25\times3.34\times10^{5}$ .
The energy needed is $JQ=8.35\times10^{4}\ \text{J}$ .
The temperature of the water is still $∘C0\,^\circ\text{C}$ after this transfer, because all of the energy went into the change of state.

Example

Finding a specific latent heat

A $W60\ \text{W}$ heater is switched on for $s500\ \text{s}$ in a funnel of ice at $∘C0\,^\circ\text{C}$ , and $kg0.090\ \text{kg}$ of water is collected.
The energy supplied is $JE=P\times t=60\times500=30\,000\ \text{J}$ .
Rearranging gives $0000.090L=\dfrac{Q}{m}=\dfrac{30\,000}{0.090}$ .
The specific latent heat of fusion works out as $J/kgL=3.3\times10^{5}\ \text{J/kg}$ to two significant figures.
Some ice would melt from the warmth of the room as well, which is why a second funnel of ice with no heater is used as a control.

Example

Heating water and then boiling it

A mass of $kg0.10\ \text{kg}$ of water at $∘C80\,^\circ\text{C}$ is heated to $∘C100\,^\circ\text{C}$ and then boiled away completely.
The first stage is a temperature change, so $J\Delta Q=m\,c\,\Delta\theta=0.10\times4200\times20=8400\ \text{J}$ .
The second stage is a change of state, so $JQ=m\,L=0.10\times2.26\times10^{6}=2.26\times10^{5}\ \text{J}$ .
The total energy is $J8400+2.26\times10^{5}=2.34\times10^{5}\ \text{J}$ .
Boiling the water accounts for about $96%96\%$ of the total, even though it produces no change in temperature.

Exam technique

Setting out the calculation

Write $LQ=m\,L$ before substituting, then rearrange on its own line when the question asks for the mass or for $L$ .
State which specific latent heat you are using, because a question involving both melting and boiling gives two values to choose between.
Label the two stages of a combined problem as separate calculations, so partial credit survives an arithmetic slip in one of them.
Give the answer in standard form with a unit, since latent heat answers run into hundreds of thousands of joules.
Name the energy escaping to the surroundings and say that it makes the value too high when a question asks you to comment on an experimental result.

Common Mistake

Do not include a temperature change in this equation, because $LQ=m\,L$ has no temperature term.
Do not use the specific latent heat of fusion for boiling, because the vaporisation value is far larger.
Do not use the total mass when only part of the substance changes state.
Do not answer a two stage question with a single calculation, because the marks are split between the stages.
Do not forget to convert the mass into kilograms, because the specific latent heat is quoted per kilogram.

Self review

State the latent heat equation and give the unit of each quantity in it.
Calculate the energy released when $kg0.40\ \text{kg}$ of water at $∘C0\,^\circ\text{C}$ freezes.
Rearrange the equation to find the mass melted by a known amount of energy.
Explain why the equation contains no temperature term.
Describe how you would find the total energy needed to melt ice and then warm the water produced.
Explain why an experimental value of specific latent heat usually comes out too high.

14.2.4 Specific latent heat calculation

The latent heat equation

Rearranging and supplying the energy

Problems with more than one stage

Energy needed to melt ice

Finding a specific latent heat

Heating water and then boiling it

Setting out the calculation

Review 14.2.4 Specific latent heat calculation by teaching Genie

14.2.4 Specific latent heat calculation

The latent heat equation

Rearranging and supplying the energy

Problems with more than one stage

Energy needed to melt ice

Finding a specific latent heat

Heating water and then boiling it

Setting out the calculation

Review 14.2.4 Specific latent heat calculation by teaching Genie

14.2.4 Specific latent heat calculation

The latent heat equation

Rearranging and supplying the energy

Problems with more than one stage

Energy needed to melt ice

Finding a specific latent heat

Heating water and then boiling it

Setting out the calculation

Review 14.2.4 Specific latent heat calculation by teaching Genie

14.2.4 Specific latent heat calculation Revision Guide

14.2.4 Specific latent heat calculation

The latent heat equation

Rearranging and supplying the energy

Problems with more than one stage

Energy needed to melt ice

Finding a specific latent heat

Heating water and then boiling it

Setting out the calculation

Review 14.2.4 Specific latent heat calculation by teaching Genie

14.2.4 Specific latent heat calculation Revision Guide