Specific Heat Capacity
Key idea: Define and use heat capacity and specific heat capacity, solve energy-balance problems, and evaluate an electrical heating experiment.
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The core idea
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Learning objectives
- Define and use heat capacity and specific heat capacity in energy balances.
- Define and use specific latent heat in phase-change energy balances.
1. Definitions (Must Know)
- Heat capacity, C (J K⁻¹), is the energy required to raise the temperature of an object by 1 K: C = Q/(Δ T)
- Specific heat capacity, c (J kg⁻¹ K⁻¹), is the energy required to raise the temperature of 1 kg of a substance by 1 K: c = Q/(mΔ T)
Therefore C = mc and, when c is effectively constant over the interval, Q = mcΔ T
Here Q is the energy transferred by heating, m is mass and Δ T = T_f-Tᵢ is the signed temperature change.
2. Key Ideas (What Earns Marks)
- A temperature difference has the same numerical value in kelvin and degrees Celsius.
- Heating an object gives Q > 0 and Δ T > 0; cooling gives both quantities negative under a signed convention.
- In an isolated calorimetry model: energy lost by hotter parts = energy gained by cooler parts
- Include the container’s heat capacity when it is not negligible.
- Q = mcΔ T describes a temperature change within one state. During a phase change, use specific latent heat instead.
A calorimetry question may involve the sample, water, calorimeter and surroundings. List every part that changes temperature before writing the energy balance.
3. Detailed Explanations
A. Energy balance at thermal equilibrium
For two bodies placed in an insulated container, conservation of energy gives mₕcₕ(Tₕ-T_f) = m_cc_c(T_f-T_c) where the left side is the positive energy lost by the hotter body and the right side is the positive energy gained by the cooler body.
This magnitude form avoids sign errors. If you use signed changes instead, write ∑ Q = 0 consistently.
B. Electrical method
If a heater has potential difference Vₑ, current I and heating time t, its electrical input is Eᵢₙ = VₑIt The ideal model sets VₑIt = mcΔ T. In practice, VₑIt = mcΔ T + Eₐₚₚₐᵣₐₜᵤₛ + Eₗₒₛₛ so insulation, a lid and good thermal contact reduce systematic error.
4. Common Mistakes
- Using the object’s total heat capacity C where the question gives specific heat capacity c.
- Omitting the mass or using grams instead of kilograms.
- Treating temperature as energy; temperature is related to mean microscopic kinetic energy, not the total energy stored.
- Applying mcΔ T across melting or boiling.
5. Exam Tips
- Write the energy balance in words before substituting.
- State the isolation assumption if energy loss is neglected.
- Check units: kg × J kg⁻¹ K⁻¹ × K = J.
- A final equilibrium temperature must lie between the initial temperatures unless another energy transfer or phase change occurs.
6. Worked Examples
Modelled example 1
Heating a metal block
Problem
Study the worked solution
Find the temperature change
Method
The block warms by 45.0 K.Reason
Kelvin and Celsius temperature intervals have the same numerical size.Working
Δ T = 65.0-20.0 = 45.0 KApply the material relation
Method
Use Q = mcΔ T.Reason
The block remains in one phase and its specific heat capacity is treated as constant.Working
Q = (0.80)(450)(45.0)Evaluate
Method
The energy transferred is 1.62 × 10⁴ J.Reason
The mass, specific heat capacity and temperature interval combine to joules.Working
Q = 1.62 × 10⁴ J
Common misconception 2
Mixing two samples of water
Learner claim
Try this before viewing the solution
View solution step by step
Reject the simple average
Method
The equilibrium temperature is not the unweighted midpoint.Reason
The cooler sample has greater mass and therefore greater heat capacity.Working
0.30 kg > 0.20 kgWrite the isolated balance
Method
Energy lost by the hot water equals energy gained by the cool water.Reason
Both are water, so the common specific heat capacity cancels.Working
0.20(80-T_f) = 0.30(T_f-20)Solve and check
Method
The equilibrium temperature is 44°C.Reason
It lies between the initial temperatures and closer to 20°C because more cool water is present.Working
16-0.20T_f = 0.30T_f-6 ⇒ T_f = 44°C
7. Mind Stretchers
Mind stretcher 1: Direction of experimental biasExtension
In the electrical method, a student assumes all electrical energy heats the block although some escapes. Is the calculated c too high or too low?
Show Answer
The student uses c = Eᵢₙ/(mΔ T). Because Eᵢₙ is greater than the energy actually stored in the block for the measured Δ T, the calculated c is too high.
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Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027