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D2.10 · Solve heat, temperature-change, and phase-change problems

Learn to solve heat, temperature-change, and phase-change problems through clear examples and targeted practice.

Ontario Grade 11 Physics

Energy and Society

Using energy, specific heat capacity, and latent heat

Heat problems describe energy moving into or out of a chosen system. First decide what the system is. For example, the system might be one piece of metal, a sample of ice, or warm water together with ice. Heat is energy transferred because of a temperature difference. It is measured in joules (J). Temperature describes how hot or cold something is. It is a scalar: it has a value but no direction. Heat transfer has a direction, so state whether energy enters or leaves the system. In this lesson, heat entering the chosen system is positive and heat leaving it is negative.

What you will learn

1. Prerequisite bridge: temperature and energy

A temperature change is the final temperature minus the initial temperature. Temperatures used in these calculations can be in degrees Celsius because a change of one degree Celsius has the same size as a change of one kelvin. The numerical temperature change is therefore the same in either scale. Use the units shown in the question and keep them consistent.
Mass is the amount of matter in a sample. In these equations, use mass in kilograms (kg). If a mass is given in grams, divide by 1000 to convert it to kilograms. Energy is measured in joules. Specific heat capacity tells how much energy is needed to raise one kilogram of a substance by one degree Celsius. Its unit is joules per kilogram per degree Celsius.
A substance can also change state. A phase change is a change between solid, liquid, and gas. During a phase change, energy can enter or leave while the temperature stays constant. For example, ice at its melting point can absorb energy as it melts without becoming warmer until the melting is complete.
ΔT=Tf−Ti\Delta T=T_f-T_i

2. Temperature changes: specific heat capacity

Use the specific-heat equation when a substance changes temperature but does not change state during the part of the problem being calculated. The symbol QQ means heat transferred to the chosen system. The symbol mm is its mass, and cc is its specific heat capacity. A positive value of QQ means energy enters the system. A negative value means energy leaves it.
The sign follows from the temperature change. If a sample warms, its final temperature is higher than its initial temperature, so ΔT\Delta T and QQ are positive. If it cools, both are negative. The size of QQ depends on the mass, the substance, and the size of the temperature change.
Before substituting, identify the system, the known values, and the unknown. State the direction of energy transfer in words. Keep units in the substitution so you can check that the result is in joules. Do not use this equation for the energy absorbed or released during a phase change.
Q=mcΔTQ=mc\Delta T

3. Phase changes and combined energy problems

During a phase change, use latent heat rather than specific heat. Latent heat of fusion applies to melting or freezing. Latent heat of vaporization applies to boiling or condensing. The symbol LL stands for the relevant latent heat per kilogram, in joules per kilogram. For a specified mass, the transferred energy is the mass multiplied by the relevant latent heat.
With the sign convention used here, melting and vaporizing require energy to enter the substance, so QQ is positive. Freezing and condensing release energy from the substance, so QQ is negative. The temperature stays constant during the phase change itself. If a problem includes both a temperature change and a phase change, calculate the energy for each part separately, then combine the energy amounts.
For two substances exchanging energy in an insulated situation, energy lost by one is gained by the other. Insulated means that, for the model in the problem, energy does not escape to the surroundings. Use this model only when the question supports it. A useful plan is to list each stage, choose the correct equation for that stage, and keep track of which substance gains or loses energy.
Q=mLQ=mL

4. A reliable solving routine

Start by naming the system and defining the positive direction for energy transfer: into the system is positive. List the known values and the unknown. Identify whether each stage changes temperature, changes phase, or does both in sequence. Then select the matching equation.
Substitute values with units. Carry out the arithmetic and round the final result to a sensible number of significant figures based on the given values. Include the unit and state the direction in words, such as “energy enters the sample” or “the water loses energy.”
Finally, check whether the answer makes sense. A warming sample should have positive heat transferred into it. A cooling sample should have negative heat transferred to it. A phase-change energy should be positive for melting and negative for freezing under this sign convention. Check that an energy answer is in joules and that its size is plausible for the sample and temperature change.

Worked example

1. Heating a metal sample

A 0.250 kg aluminum sample warms from 20.0 °C to 85.0 °C. Use a specific heat capacity of 900 J/(kg·°C). Find the heat transferred to the aluminum.
  1. Define the system
    The system is the aluminum sample. Heat entering it is positive. The sample warms, so energy transfers into it. The unknown is the heat transferred, QQ.
  2. Find the temperature change
    Subtract the initial temperature from the final temperature. A positive result agrees with the stated warming.
    ΔT=85.0 ∘C−20.0 ∘C=65.0 ∘C\Delta T=85.0\ ^\circ\mathrm{C}-20.0\ ^\circ\mathrm{C}=65.0\ ^\circ\mathrm{C}
  3. Use specific heat
    There is a temperature change and no phase change, so use the specific-heat relationship. Substitute the mass, specific heat capacity, and temperature change with their units.
    Q=(0.250 kg)(900 J/(kg⋅∘C))(65.0 ∘C)=14625 JQ=(0.250\ \mathrm{kg})(900\ \mathrm{J/(kg\cdot{}^\circ C)})(65.0\ ^\circ\mathrm{C})=14625\ \mathrm{J}
  4. Round and check
    The mass has three significant figures, while the supplied specific heat has two. Report two significant figures. Kilograms and degrees Celsius cancel, leaving joules. The positive result agrees with energy entering a warming sample.
    Q=1.5×104 JQ=1.5\times10^4\ \mathrm{J}
Answer: The aluminum gains 1.5×1041.5\times10^4 J of energy.
Check: The sign is positive because the sample warms. The unit is joules, and the energy is a reasonable amount for a 0.250 kg sample warmed by 65.0 °C.

Worked example

2. Melting ice

A 0.0800 kg piece of ice is already at its melting temperature. How much energy is needed to melt it? Use a latent heat of fusion of 3.34×1053.34\times10^5 J/kg.
  1. Define the system
    The system is the ice. Energy entering it is positive. The ice melts, so its state changes but its temperature does not change during the melting stage. The unknown is the heat absorbed.
  2. Choose the phase-change relationship
    Because the problem describes melting at the melting temperature, use latent heat of fusion rather than the specific-heat equation.
    Q=mLfQ=mL_f
  3. Substitute and calculate
    Use the ice mass in kilograms and the latent heat of fusion in joules per kilogram. The kilogram units cancel, leaving joules.
    Q=(0.0800 kg)(3.34×105 J/kg)=26720 JQ=(0.0800\ \mathrm{kg})(3.34\times10^5\ \mathrm{J/kg})=26720\ \mathrm{J}
  4. Round and check
    Both given values have three significant figures. Melting requires energy to enter the ice, so the answer is positive. The result has units of joules.
    Q=2.67×104 JQ=2.67\times10^4\ \mathrm{J}
Answer: The ice absorbs 2.67×1042.67\times10^4 J to melt.
Check: A positive energy amount is appropriate for melting. The calculation uses mass times energy per kilogram, giving joules. The temperature remains constant during the phase change.

Worked example

3. Warm water melting ice, then cooling

In an insulated container, 0.200 kg of water at 60.0 °C is mixed with 0.0500 kg of ice at its melting temperature. Assume all the ice melts and the final mixture is liquid water. Use c=4180c=4180 J/(kg·°C) for liquid water and Lf=3.34×105L_f=3.34\times10^5 J/kg. Find the final temperature.
  1. Define the system and energy directions
    The system is the warm water and the ice together. Energy does not leave this system in the insulated model. The warm water loses energy as it cools; the ice gains energy to melt and then the melted ice warms. The unknown is the final temperature, TfT_f.
  2. Calculate energy available from the warm water
    First find the energy the original water would lose in cooling from 60.0 °C to 0.0 °C. This gives the energy available for melting and warming the ice. The magnitude is positive here because it is being compared as an amount of energy released.
    Qreleased=(0.200 kg)(4180 J/(kg⋅∘C))(60.0 ∘C)=50160 JQ_{\mathrm{released}}=(0.200\ \mathrm{kg})(4180\ \mathrm{J/(kg\cdot{}^\circ C)})(60.0\ ^\circ\mathrm{C})=50160\ \mathrm{J}
  3. Find energy used to melt the ice
    The ice starts at its melting temperature, so use latent heat for this stage. Subtract this energy from the amount released by the warm water to find the energy left to warm all the liquid water.
    Qmelt=(0.0500 kg)(3.34×105 J/kg)=16700 JQ_{\mathrm{melt}}=(0.0500\ \mathrm{kg})(3.34\times10^5\ \mathrm{J/kg})=16700\ \mathrm{J}
  4. Find the final temperature
    After melting, the total liquid-water mass is 0.250 kg. The remaining energy warms this water from 0.0 °C to the final temperature. Equating the remaining energy to the specific-heat energy gives the result.
    Tf=50160 J−16700 J(0.250 kg)(4180 J/(kg⋅∘C))=32.0 ∘CT_f=\frac{50160\ \mathrm{J}-16700\ \mathrm{J}}{(0.250\ \mathrm{kg})(4180\ \mathrm{J/(kg\cdot{}^\circ C)})}=32.0\ ^\circ\mathrm{C}
Answer: The final temperature is 32.0 °C.
Check: The warm water releases more energy than the ice needs to melt, so all the ice can melt. The remaining energy produces a final temperature above 0.0 °C and below 60.0 °C, as expected. Energy is conserved within the insulated system.

Common mistakes and how to avoid them

Using the specific-heat equation for melting or boiling.
Correction: Use latent heat for energy during a phase change. Use specific heat for a temperature change within one state.
Assuming the temperature rises while a pure substance is changing phase.
Correction: During the phase change itself, energy changes the state while the temperature remains constant.
Using a negative temperature change for cooling but reporting a positive heat transfer to the system.
Correction: With heat into the system defined as positive, cooling gives negative ΔT\Delta T and negative QQ.
Leaving mass in grams when using specific heat or latent heat values given per kilogram.
Correction: Convert grams to kilograms before substituting, or convert the material constant consistently.
Adding the energy amounts in an insulated exchange without checking which substance loses energy.
Correction: Track energy direction. Energy lost by one part of the system is gained by another in the insulated model.

Lesson summary

Check your understanding

Question 1

A sample cools from 48.0 °C to 18.0 °C. With heat into the sample defined as positive, what is the sign of its heat transfer?
  1. Positive, because its temperature changes.
  2. Negative, because energy leaves the sample as it cools.
  3. Zero, because cooling does not involve heat transfer.
  4. correctIndex: 1
Show answer and explanation
Negative, because energy leaves the sample as it cools.
Cooling means the sample loses energy. Under the stated convention, heat leaving the system is negative.

Question 2

Which relationship is appropriate for finding the energy needed to freeze liquid at its freezing temperature?
  1. Q=mcΔTQ=mc\Delta T
  2. Q=mLfQ=mL_f
  3. Q=mcQ=mc
  4. correctIndex: 1
Show answer and explanation
Q=mLfQ=mL_f
Freezing is a phase change. Use latent heat of fusion. With heat into the liquid-water system defined as positive, freezing has negative heat transfer.

Question 3

A substance absorbs energy while melting at its melting temperature. What happens to its temperature during the melting stage?
  1. It stays constant during the phase change.
  2. It rises in direct proportion to the absorbed energy.
  3. It falls because the substance absorbs energy.
  4. correctIndex: 0
Show answer and explanation
It stays constant during the phase change.
During melting, transferred energy changes the state. The temperature remains constant until the phase change is complete.

Key terms

System
The object or group of objects chosen for a physics calculation.
Heat
Energy transferred into or out of a system because of a temperature difference.
Specific heat capacity
The energy needed to raise the temperature of one kilogram of a substance by one degree Celsius.
Phase change
A change between solid, liquid, and gas.
Latent heat
Energy transferred per kilogram during a phase change, without a temperature change during that change.
Insulated
Modelled so that energy does not transfer between the system and its surroundings.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation D2.10. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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