Specific Heat Capacity

Key idea: Define and use heat capacity and specific heat capacity, solve energy-balance problems, and evaluate an electrical heating experiment.

  • GCE A-Level H2 Physics 2027
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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.
Choose the complete thermal system

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

Specific heat capacity experiment setupAn insulated metal block containing an electric heater and thermometer. An ammeter is in series with the heater and a voltmeter is connected across it, while a stopwatch measures heating time.Electrical inputd.c. powersupplyAMetal blockInsulation around blockHeaterThermometeror probeVacross heaterStopwatchRecord V, I, t, m and Δθc = VIt / (mΔθ)
Measure potential difference, current, time, mass and temperature rise. The ideal estimate c = VIt/(mΔT) is too large if the measured electrical input includes energy that warms the apparatus or escapes to the surroundings.

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

Core

Problem

A 0.80 kg block with c = 450 J kg⁻¹ K⁻¹ is warmed from 20.0°C to 65.0°C. Find the energy transferred.
Study the worked solution
  1. 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 K
  2. Apply 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)
  3. 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

Find and correct the mistake

Learner claim

0.20 kg of water at 80°C is mixed with 0.30 kg at 20°C in an insulated container. A learner averages the two temperatures and predicts 50°C. Diagnose the claim and find the equilibrium temperature.

Try this before viewing the solution

Unit: °C

View solution step by step
  1. 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 kg
  2. Write 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)
  3. 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