Heat Capacity & Specific Heat Capacity

Key idea: Learn heat capacity and specific heat capacity, use Q=mcΔθ and Q=Pt for calculations, and avoid common unit mistakes for O Level Physics.

  • SEC G3 Physics 2027
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Learning objectives

  • Compare physical properties of solids, liquids and gases
  • Explain state properties using particle arrangement, motion, forces and separation
  • Infer random molecular motion from a Brownian-motion experiment
  • Relate temperature rise to increased average kinetic energy of particles
  • Explain gas pressure using particle collisions with container walls
  • Explain heating from higher to lower temperature until thermal equilibrium
  • Describe conduction in solids through particle vibration and mobile electrons
  • Describe convection in fluids through density changes and bulk motion
  • Explain that energy transfer by electromagnetic radiation needs no material medium
  • Explain how surface colour, texture, temperature and area affect radiation transfer rate
  • Apply conduction, convection and radiation in everyday systems
  • describe internal energy as an energy store that is made up of the total kinetic energy associated with the random motion of the particles and the total potential energy between the particles in the system
  • Define heat capacity and specific heat capacity
  • Apply energy transferred = mass × specific heat capacity × temperature change
  • describe melting/solidification and boiling/condensation as processes of energy transfer without a change in temperature
  • Explain the difference between boiling and evaporation
  • Define latent heat and specific latent heat
  • Apply energy transferred for a change of state = mass × specific latent heat
  • Explain latent heat using particle behaviour
  • Sketch and interpret a cooling curve

1. Definition

A. Heat capacity

Heat capacity, C, of a body is the energy required to raise its temperature by 1 K (or 1°C), without change of state.

C = Q/(Δ θ)

B. Specific heat capacity

Specific heat capacity, c, of a substance is the energy required to raise the temperature of 1 kg of the substance by 1 K (or 1°C), without change of state.

Q = mcΔ θ

2. Key Ideas

  • Units:
    • C in J K⁻¹ (or J °C⁻¹)
    • c in J kg⁻¹ K⁻¹ (or J kg⁻¹ °C⁻¹)
    • Q in J, m in kg, Δ θ in K or °C
  • Temperature changes: 1 K is the same size as 1°C, so you can use Δ θ in either unit.
  • Rearrangements:
    • Q = CΔ θ
    • c = Q/(mΔ θ)
    • C = mc
  • For the same energy input Q:
    • larger m or larger c → smaller temperature rise (Δ θ).
Substance (approx.)Specific heat capacity, c (J kg⁻¹ K⁻¹)
water4200
aluminium900
iron450
copper390
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 heater potential difference and current, heating time, sample mass and temperature rise. Insulation and good thermal contact reduce systematic error.
Specific heat capacity varies by materialComparison of typical specific heat capacity values for water and common metals.Specific heat capacity varies by materialMaterialSpecific heat capacity (J kg⁻¹ K⁻¹)
Water has a much higher specific heat capacity than many metals, so for the same mass and same energy input, water’s temperature rises less.
Data table
Materialc
Water4200
Aluminium900
Iron450
Copper390

3. Detailed Explanations

A. Heat capacity vs specific heat capacity

  • Heat capacity C applies to a particular object (depends on its mass and material).
  • Specific heat capacity c is a property of the material (per kilogram).

They are linked by:

C = mc

So a larger mass of the same material has a larger heat capacity.

B. What does “high specific heat capacity” mean?

If a material has a high c:

  • it needs a lot of energy to raise its temperature by 1°C per kg
  • for the same heating power, it warms up more slowly
  • under otherwise identical cooling conditions, its temperature changes more slowly

C. Measuring the specific heat capacity of a metal block

  1. Measure the block’s mass m with a balance.
  2. Insert the heater and thermometer into their holes. Add a little thermal-contact material if instructed, so air gaps do not slow energy transfer.
  3. Wrap the block in insulation and record its initial temperature.
  4. Connect an ammeter in series with the heater and a voltmeter across it. Switch on and record I, V and heating time t.
  5. Record the final temperature and calculate Δθ.

The electrical energy supplied is E = VIt (or E = Pt if power is measured). Under the ideal assumption that all of it raises the block’s internal energy,

c = VIt/mΔθ.

Why the experimental value can be too large

Some electrical energy warms the heater and thermometer or is transferred to the surroundings. Using the full VIt as though it all heated the block therefore tends to overestimate c. Insulate the block, ensure good thermal contact, and use a sufficiently large temperature rise to reduce the percentage effect of these transfers.

D. A stronger data method

Record temperature at regular times while V and I are approximately constant, then plot temperature against time. The early gradient can be used with VI ≈ mc(Δθ/Δ t). Repeated readings reveal anomalies and make the trend more reliable than a single start–finish pair, although heat loss still needs evaluation.

4. Common Mistakes

  • Using mass in g instead of kg without converting.
  • Using the absolute temperature instead of the temperature change (use Δ θ).
  • Forgetting the “no change of state” condition (during melting/boiling, this formula is not used).
  • Mixing up C (heat capacity) with c (specific heat capacity).
  • Writing units incorrectly (e.g. missing kg⁻¹ for c).
  • Claiming insulation removes all heat loss. It reduces unwanted transfer; it does not make it zero.
  • Putting the voltmeter in series or the ammeter across the heater.

5. Exam Tips

  • Write the equation first, then substitute values with units:
    • Q = mcΔθ or C = Q/Δθ
  • Convert mass to kg and temperature change to K or °C.
  • If the question gives power and time, use:
    • Q = Pt
  • If the question mentions “assume no heat loss”, state it explicitly before using Q = Pt.
  • In an evaluation question, name the energy destination and its effect on the result; “heat loss causes error” is too vague.

6. Worked Examples

Modelled example 1

Heat capacity scales with mass

Core

Problem

100 g of water needs 12 600 J to warm from 30°C to 60°C. Find its heat capacity, then the heat capacity of 1000 g and the energy needed to warm that larger mass by 10°C.
Study the worked solution
  1. Find the first heat capacity

    Method

    Divide energy by temperature change.

    Reason

    Heat capacity belongs to the stated body and follows C = Q/Δθ.

    Working

    Δθ = 60-30 = 30°C, C = (12 600)/30 = 420 J K⁻¹
  2. Scale with mass

    Method

    Multiply heat capacity by ten for ten times the same substance.

    Reason

    C = mc, so heat capacity is proportional to mass when material is fixed.

    Working

    C_1000g = 10(420) = 4200 J K⁻¹
  3. Find the new energy transfer

    Method

    Multiply the larger heat capacity by its 10 K rise.

    Reason

    Q = CΔθ.

    Working

    Q = (4200)(10) = 4.2 × 10⁴ J

Guided practice 2

Specific heat capacity (direct use)

About 5 min

Problem

A 5.0 kg mass receives 20 000 J and warms from 15°C to 25°C. Find its specific heat capacity.

Use temperature change, not final temperature

Unit: J kg^-1 K^-1

Hints

Hint 1: find the rise
Δθ = 25-15.
Hint 2: rearrange before substituting
Use c = Q/(mΔθ).
View solution step by step
  1. Calculate temperature change

    Method

    Subtract initial temperature from final temperature.

    Reason

    The equation uses a change, not an absolute reading.

    Working

    Δθ = 25-15 = 10 K
  2. Calculate specific heat capacity

    Method

    Divide energy by mass and temperature change.

    Reason

    This isolates the per-kilogram, per-kelvin material property.

    Working

    c = (20 000)/(5.0)(10) = 400 J kg⁻¹ K⁻¹

Common misconception 3

Find the mass heated

Find and correct the mistake

Learner response

18 900 J warms water by 15°C, with c = 4200 J kg⁻¹ K⁻¹. A student multiplies Q, c and Δθ to find mass. Locate the error and calculate the mass.

Make mass the subject first

Unit: kg

View solution step by step
  1. Rearrange for mass

    Method

    Divide energy by specific heat capacity and temperature change.

    Reason

    Mass multiplies both factors in Q = mcΔθ.

    Working

    m = Q/cΔθ
  2. Substitute and calculate

    Method

    Use the supplied consistent SI units.

    Reason

    The specific heat-capacity unit already expects kilograms.

    Working

    m = (18 900)/(4200)(15) = 0.30 kg

Examiner practice 4

Find specific heat capacity using Q = Pt

5 marks

Examination question

A 50 W heater warms a 0.60 kg metal block from 20°C to 45°C in 90 s. Find its specific heat capacity and state one assumption. [5 marks]

Show energy, temperature change, calculation and assumption

View solution step by step
  1. Find supplied energy

    1 mark

    Method

    Multiply power by heating time.

    Reason

    One watt is one joule per second.

    Working

    Q = Pt = (50)(90) = 4500 J
  2. Find temperature change

    1 mark

    Method

    Subtract the initial reading.

    Reason

    Q = mcΔθ uses temperature rise.

    Working

    Δθ = 45-20 = 25 K
  3. Calculate and qualify the result

    3 marks

    Method

    Calculate c and assume all supplied energy heats the block.

    Reason

    Energy transferred to the surroundings or apparatus would invalidate the ideal equality.

    Working

    c = 4500/(0.60)(25) = 300 J kg⁻¹ K⁻¹

Challenge 5

Temperature rise from power

Minimal support

Changed-unknown transfer

A 300 W heater warms 0.50 kg of water for 4.0 min. Assuming no unwanted energy transfer and using c = 4200 J kg⁻¹ K⁻¹, find the temperature rise.

Convert time before combining equations

Hints

Hint 1: match watts to seconds
Convert 4.0 min to 240 s.
Hint 2: combine energy equations
Set Pt = mcΔθ.
View solution step by step
  1. Calculate heater energy

    Method

    Convert minutes to seconds and multiply by power.

    Reason

    Watts multiplied by seconds gives joules.

    Working

    Q = Pt = 300(4.0 × 60) = 72 000 J
  2. Find temperature rise

    Method

    Divide energy by mass and specific heat capacity.

    Reason

    Rearranging Q = mcΔθ isolates the unknown rise.

    Working

    Δθ = (72 000)/(0.50)(4200) ≈ 34 K

7. Mind Stretchers

Mind stretcher 1: Same energy, different temperature riseExtension

Two blocks have the same mass. Block A has a larger specific heat capacity than block B. The same amount of energy is supplied to both blocks. Which block has the larger temperature rise? Explain.

Show Answer

Block B has the larger temperature rise.

From Q = mcΔθ, for the same Q and m: Δθ = Q/mc

Larger c gives smaller Δθ.

Mind stretcher 2: Cooling downExtension

Two cups contain the same mass of water. Cup 1 cools from 60°C to 40°C. Cup 2 cools from 30°C to 10°C. Which cup transfers more energy to the surroundings? Explain.

Show Answer

They transfer the same amount.

Both have the same mass m, same c (water), and the same temperature change magnitude Δθ = 20°C, so: Q = mcΔθ

has the same magnitude for both.

8. Practice and next step

Check mass conversion, Δθ and units in the Thermal Physics quiz, then show full working in structured practice. Continue to melting and solidification, where energy can transfer without a temperature change.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027