Energy and changes of state

Definition

Latent heat

The energy transferred to or from a substance during a change of state while its temperature stays constant.

Definition

Specific latent heat

The energy needed to change the state of $kg1\ \text{kg}$ of a substance without any change in its temperature.

While a substance changes state, energy is transferred to or from it but its temperature stays the same.
This energy is called latent heat, from a Latin word meaning hidden, because a thermometer shows no sign of it.
When a substance melts or boils, it takes in energy, and that energy goes into the potential energy stores of the particles as they move further apart.
When a substance freezes or condenses, it gives out the same amount of energy as the particles move closer together again.
The energy involved depends on the mass that changes state and on the specific latent heat of the substance.
Twice the mass of ice needs twice the energy to melt, which is why a large bag of ice keeps a cool box cold for longer than a handful of cubes.

The equation

Definition

Specific latent heat of fusion

The energy needed to change $kg1\ \text{kg}$ of a substance from solid to liquid without any change in its temperature.

Definition

Specific latent heat of vaporisation

The energy needed to change $kg1\ \text{kg}$ of a substance from liquid to gas without any change in its temperature.

The equation is $heat\text{thermal energy for a change in state} = \text{mass} \times \text{specific latent heat}$ .
In symbols, this is $E=m LE = m\,L$ $E = m L$ .
1. $E$ is the energy transferred in joules, $J\text{J}$ .
2. $m$ is the mass that changes state in kilograms, $kg\text{kg}$ .
3. $L$ is the specific latent heat in joules per kilogram, $J/kg\text{J/kg}$ .
Use the specific latent heat of fusion for melting or freezing, and the specific latent heat of vaporisation for boiling or condensing.
There is no temperature in the equation, because the temperature does not change while the state is changing.
The mass must be the mass that changes state, so if only part of a block of ice melts, you use only the mass that melts.
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.
Rearranging gives $\dfrac{E}{m}$ for the specific latent heat and $\dfrac{E}{L}$ for the mass.

Calculating the energy

Write the equation, convert any mass in grams into kilograms, substitute the values, and give the answer with its unit.
Values of specific latent heat are large, so they are usually given in standard form, such as $3.34 \times 10^{5}\ \text{J/kg}$ .
Enter standard form into your calculator with the $×10x\times 10^{x}$ key rather than typing out the zeros, which avoids slipping a digit.
Some questions combine both kinds of energy change, with a substance first warming up to its melting or boiling point and then changing state.
For these, work out each stage separately, using $Δθ\Delta E = m\,c\,\Delta\theta$ for warming and $m\,L$ for changing state, and then add the energies together.

Example

Melting ice

A $kg0.25\ \text{kg}$ block of ice at $∘C0\ ^\circ\text{C}$ melts into water at $∘C0\ ^\circ\text{C}$ , and the specific latent heat of fusion of water is $3.34 \times 10^{5}\ \text{J/kg}$ .

m\,L

\times 3.34 \times 10^{5}

83\,500\ \text{J} = 8.35 \times 10^{4}\ \text{J}

Melting the ice takes $J83\,500\ \text{J}$ , and the temperature stays at $∘C0\ ^\circ\text{C}$ throughout.

Example

Finding a specific latent heat

A heater transfers $J45\,000\ \text{J}$ to water that is already boiling, and during that time $g20\ \text{g}$ of the water turns into steam.
Convert the mass into kilograms.

\dfrac{20}{1000} = 0.020\ \text{kg}

Rearrange the equation, substitute the values and calculate.

\dfrac{E}{m} = \dfrac{45\,000}{0.020} = 2.25 \times 10^{6}\ \text{J/kg}

This is the specific latent heat of vaporisation of water, close to the accepted value of $2.26 \times 10^{6}\ \text{J/kg}$ .

Example

Warming ice and then melting it

A $kg0.50\ \text{kg}$ block of ice from a freezer is at $∘C-18\ ^\circ\text{C}$ and is warmed until it has all melted at $∘C0\ ^\circ\text{C}$ .
The specific heat capacity of ice is $∘C2100\ \text{J/kg}\,^\circ\text{C}$ , and its specific latent heat of fusion is $3.34 \times 10^{5}\ \text{J/kg}$ .
Stage 1 warms the solid ice from $∘C-18\ ^\circ\text{C}$ to $∘C0\ ^\circ\text{C}$ , a temperature change of $∘C18\ ^\circ\text{C}$ .

\Delta E = m\,c\,\Delta\theta = 0.50 \times 2100 \times 18 = 18\,900\ \text{J}

Stage 2 melts the ice at a constant $∘C0\ ^\circ\text{C}$ .

m\,L = 0.50 \times 3.34 \times 10^{5} = 167\,000\ \text{J}

The total energy is the sum of the two stages.

E_{\text{total}} = 18\,900 + 167\,000 = 185\,900\ \text{J} \approx 1.9 \times 10^{5}\ \text{J}

Melting needs almost nine times as much energy as warming the ice, which is why ice sits at $∘C0\ ^\circ\text{C}$ for so long once it starts to melt.

Practical

Specific latent heat of vaporisation of water

Aim: to measure the energy needed to boil away a measured mass of water, and use it to find the specific latent heat of vaporisation of water.
Apparatus: a $V12\ \text{V}$ electric immersion heater, a low-voltage power supply, a joulemeter, a $cm3250\ \text{cm}^3$ glass beaker wrapped in insulating foam, an electronic balance reading to $g0.1\ \text{g}$ , a clamp stand with boss and clamp, a heatproof mat, water and eye protection.
Method:
- Stand the insulated beaker on the electronic balance and pour in about $cm3150\ \text{cm}^3$ of water.
- Clamp the heater so that its coil is fully under the water but does not touch the beaker, so that its weight does not rest on the balance.
- Connect the heater to the power supply through the joulemeter, switch on, and heat the water until it is boiling steadily.
- Once the water is boiling, reset the joulemeter to zero and record the balance reading at the same moment.
- Let the water boil for about $5$ minutes, checking that the coil stays covered the whole time.
- Record the joulemeter reading and the new balance reading at the same moment, then switch off.
- Subtract the final balance reading from the first one to find the mass of water that has turned into steam.
Why wait for boiling: the readings start only once the water is boiling, so that all of the energy recorded goes into changing state rather than into warming the water.
Maths: the specific latent heat is $\dfrac{E}{m}$ , where $E$ is the joulemeter reading in $J\text{J}$ and $m$ is the mass of water boiled away in $kg\text{kg}$ .
Results: a typical class result is about $2.4 \times 10^{6}\ \text{J/kg}$ , a little above the accepted value of $2.26 \times 10^{6}\ \text{J/kg}$ .
Alternative: the specific latent heat of fusion of stearic acid can be found in the same way, by recording the energy supplied only while its temperature stays level at about $∘C69\ ^\circ\text{C}$ and measuring the mass of solid that melts in that time.
Watch out: energy lost to the surroundings makes the calculated value too high, splashing water out of the beaker makes it too low, and the bubbling makes the balance reading flicker, so take the reading quickly and at the same moment as the joulemeter reading.
Safety: wear eye protection, keep hands away from the steam because it scalds, never let the heater coil run dry because it becomes dangerously hot, and keep the power supply away from the water.

Common Mistake

Do not include a temperature change in $m\,L$ , because the temperature does not change while the state is changing.
Do not use the total mass of a substance when only part of it changes state.
Do not use the latent heat of vaporisation in a melting question, or the latent heat of fusion in a boiling question.
Do not say that energy disappears during a change of state, because it is stored in the potential energy stores of the particles and is given out again when the substance changes back.

Exam technique

Multi-step energy questions

In a question with several stages, label each stage, such as warming the ice and melting the ice, so the examiner can follow your method.
Each stage usually carries its own marks, so a slip in one stage does not cost you the marks for the other.
Check that you have chosen the right specific latent heat by matching it to the change of state named in the question.
Give large answers in standard form or in kilojoules, and make sure the unit matches the number you write.

Self review

Write the equation that links thermal energy for a change in state, mass and specific latent heat.
Explain why the equation for a change of state does not include a change in temperature.
Calculate the energy needed to melt $kg0.40\ \text{kg}$ of ice at $∘C0\ ^\circ\text{C}$ , using $\times 10^{5}\ \text{J/kg}$ .
State why the readings in the latent heat practical start only once the water is boiling.
Describe how to find the total energy needed to warm ice from below $∘C0\ ^\circ\text{C}$ and then melt it.

Energy and changes of state

Definition

Latent heat

The energy transferred to or from a substance during a change of state while its temperature stays constant.

Definition

Specific latent heat

The energy needed to change the state of $kg1\ \text{kg}$ of a substance without any change in its temperature.

While a substance changes state, energy is transferred to or from it but its temperature stays the same.
This energy is called latent heat, from a Latin word meaning hidden, because a thermometer shows no sign of it.
When a substance melts or boils, it takes in energy, and that energy goes into the potential energy stores of the particles as they move further apart.
When a substance freezes or condenses, it gives out the same amount of energy as the particles move closer together again.
The energy involved depends on the mass that changes state and on the specific latent heat of the substance.
Twice the mass of ice needs twice the energy to melt, which is why a large bag of ice keeps a cool box cold for longer than a handful of cubes.

The equation

Definition

Specific latent heat of fusion

The energy needed to change $kg1\ \text{kg}$ of a substance from solid to liquid without any change in its temperature.

Definition

Specific latent heat of vaporisation

The energy needed to change $kg1\ \text{kg}$ of a substance from liquid to gas without any change in its temperature.

The equation is $heat\text{thermal energy for a change in state} = \text{mass} \times \text{specific latent heat}$ .
In symbols, this is $E=m LE = m\,L$ $E = m L$ .
1. $E$ is the energy transferred in joules, $J\text{J}$ .
2. $m$ is the mass that changes state in kilograms, $kg\text{kg}$ .
3. $L$ is the specific latent heat in joules per kilogram, $J/kg\text{J/kg}$ .
Use the specific latent heat of fusion for melting or freezing, and the specific latent heat of vaporisation for boiling or condensing.
There is no temperature in the equation, because the temperature does not change while the state is changing.
The mass must be the mass that changes state, so if only part of a block of ice melts, you use only the mass that melts.
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.
Rearranging gives $\dfrac{E}{m}$ for the specific latent heat and $\dfrac{E}{L}$ for the mass.

Calculating the energy

Write the equation, convert any mass in grams into kilograms, substitute the values, and give the answer with its unit.
Values of specific latent heat are large, so they are usually given in standard form, such as $3.34 \times 10^{5}\ \text{J/kg}$ .
Enter standard form into your calculator with the $×10x\times 10^{x}$ key rather than typing out the zeros, which avoids slipping a digit.
Some questions combine both kinds of energy change, with a substance first warming up to its melting or boiling point and then changing state.
For these, work out each stage separately, using $Δθ\Delta E = m\,c\,\Delta\theta$ for warming and $m\,L$ for changing state, and then add the energies together.

Example

Melting ice

A $kg0.25\ \text{kg}$ block of ice at $∘C0\ ^\circ\text{C}$ melts into water at $∘C0\ ^\circ\text{C}$ , and the specific latent heat of fusion of water is $3.34 \times 10^{5}\ \text{J/kg}$ .

m\,L

\times 3.34 \times 10^{5}

83\,500\ \text{J} = 8.35 \times 10^{4}\ \text{J}

Melting the ice takes $J83\,500\ \text{J}$ , and the temperature stays at $∘C0\ ^\circ\text{C}$ throughout.

Example

Finding a specific latent heat

A heater transfers $J45\,000\ \text{J}$ to water that is already boiling, and during that time $g20\ \text{g}$ of the water turns into steam.
Convert the mass into kilograms.

\dfrac{20}{1000} = 0.020\ \text{kg}

Rearrange the equation, substitute the values and calculate.

\dfrac{E}{m} = \dfrac{45\,000}{0.020} = 2.25 \times 10^{6}\ \text{J/kg}

This is the specific latent heat of vaporisation of water, close to the accepted value of $2.26 \times 10^{6}\ \text{J/kg}$ .

Example

Warming ice and then melting it

A $kg0.50\ \text{kg}$ block of ice from a freezer is at $∘C-18\ ^\circ\text{C}$ and is warmed until it has all melted at $∘C0\ ^\circ\text{C}$ .
The specific heat capacity of ice is $∘C2100\ \text{J/kg}\,^\circ\text{C}$ , and its specific latent heat of fusion is $3.34 \times 10^{5}\ \text{J/kg}$ .
Stage 1 warms the solid ice from $∘C-18\ ^\circ\text{C}$ to $∘C0\ ^\circ\text{C}$ , a temperature change of $∘C18\ ^\circ\text{C}$ .

\Delta E = m\,c\,\Delta\theta = 0.50 \times 2100 \times 18 = 18\,900\ \text{J}

Stage 2 melts the ice at a constant $∘C0\ ^\circ\text{C}$ .

m\,L = 0.50 \times 3.34 \times 10^{5} = 167\,000\ \text{J}

The total energy is the sum of the two stages.

E_{\text{total}} = 18\,900 + 167\,000 = 185\,900\ \text{J} \approx 1.9 \times 10^{5}\ \text{J}

Melting needs almost nine times as much energy as warming the ice, which is why ice sits at $∘C0\ ^\circ\text{C}$ for so long once it starts to melt.

Practical

Specific latent heat of vaporisation of water

Aim: to measure the energy needed to boil away a measured mass of water, and use it to find the specific latent heat of vaporisation of water.
Apparatus: a $V12\ \text{V}$ electric immersion heater, a low-voltage power supply, a joulemeter, a $cm3250\ \text{cm}^3$ glass beaker wrapped in insulating foam, an electronic balance reading to $g0.1\ \text{g}$ , a clamp stand with boss and clamp, a heatproof mat, water and eye protection.
Method:
- Stand the insulated beaker on the electronic balance and pour in about $cm3150\ \text{cm}^3$ of water.
- Clamp the heater so that its coil is fully under the water but does not touch the beaker, so that its weight does not rest on the balance.
- Connect the heater to the power supply through the joulemeter, switch on, and heat the water until it is boiling steadily.
- Once the water is boiling, reset the joulemeter to zero and record the balance reading at the same moment.
- Let the water boil for about $5$ minutes, checking that the coil stays covered the whole time.
- Record the joulemeter reading and the new balance reading at the same moment, then switch off.
- Subtract the final balance reading from the first one to find the mass of water that has turned into steam.
Why wait for boiling: the readings start only once the water is boiling, so that all of the energy recorded goes into changing state rather than into warming the water.
Maths: the specific latent heat is $\dfrac{E}{m}$ , where $E$ is the joulemeter reading in $J\text{J}$ and $m$ is the mass of water boiled away in $kg\text{kg}$ .
Results: a typical class result is about $2.4 \times 10^{6}\ \text{J/kg}$ , a little above the accepted value of $2.26 \times 10^{6}\ \text{J/kg}$ .
Alternative: the specific latent heat of fusion of stearic acid can be found in the same way, by recording the energy supplied only while its temperature stays level at about $∘C69\ ^\circ\text{C}$ and measuring the mass of solid that melts in that time.
Watch out: energy lost to the surroundings makes the calculated value too high, splashing water out of the beaker makes it too low, and the bubbling makes the balance reading flicker, so take the reading quickly and at the same moment as the joulemeter reading.
Safety: wear eye protection, keep hands away from the steam because it scalds, never let the heater coil run dry because it becomes dangerously hot, and keep the power supply away from the water.

Common Mistake

Do not include a temperature change in $m\,L$ , because the temperature does not change while the state is changing.
Do not use the total mass of a substance when only part of it changes state.
Do not use the latent heat of vaporisation in a melting question, or the latent heat of fusion in a boiling question.
Do not say that energy disappears during a change of state, because it is stored in the potential energy stores of the particles and is given out again when the substance changes back.

Exam technique

Multi-step energy questions

In a question with several stages, label each stage, such as warming the ice and melting the ice, so the examiner can follow your method.
Each stage usually carries its own marks, so a slip in one stage does not cost you the marks for the other.
Check that you have chosen the right specific latent heat by matching it to the change of state named in the question.
Give large answers in standard form or in kilojoules, and make sure the unit matches the number you write.

Self review

Write the equation that links thermal energy for a change in state, mass and specific latent heat.
Explain why the equation for a change of state does not include a change in temperature.
Calculate the energy needed to melt $kg0.40\ \text{kg}$ of ice at $∘C0\ ^\circ\text{C}$ , using $\times 10^{5}\ \text{J/kg}$ .
State why the readings in the latent heat practical start only once the water is boiling.
Describe how to find the total energy needed to warm ice from below $∘C0\ ^\circ\text{C}$ and then melt it.

1.2.6 Energy transfer during a change of state

Energy and changes of state

The equation

Calculating the energy

Melting ice

Finding a specific latent heat

Warming ice and then melting it

Specific latent heat of vaporisation of water

Multi-step energy questions

Review 1.2.6 Energy transfer during a change of state by teaching Genie

1.2.6 Energy transfer during a change of state

Energy and changes of state

The equation

Calculating the energy

Melting ice

Finding a specific latent heat

Warming ice and then melting it

Specific latent heat of vaporisation of water

Multi-step energy questions

Review 1.2.6 Energy transfer during a change of state by teaching Genie

1.2.6 Energy transfer during a change of state

Energy and changes of state

The equation

Calculating the energy

Melting ice

Finding a specific latent heat

Warming ice and then melting it

Specific latent heat of vaporisation of water

Multi-step energy questions

Review 1.2.6 Energy transfer during a change of state by teaching Genie

1.2.6 Energy transfer during a change of state Revision Guide

1.2.6 Energy transfer during a change of state

Energy and changes of state

The equation

Calculating the energy

Melting ice

Finding a specific latent heat

Warming ice and then melting it

Specific latent heat of vaporisation of water

Multi-step energy questions

Review 1.2.6 Energy transfer during a change of state by teaching Genie

1.2.6 Energy transfer during a change of state Revision Guide