Mini Physics
Part of Mini EducationSwitch subject

Heat Capacity and Specific Heat Capacity

Distinguish an object’s heat capacity from a material’s specific heat capacity, and calculate energy transfers and temperature changes.

  • SEC G3 Physics 2027

Reading settings

On this page

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 transferred to the sample, with no change of state and approximately constant c:
    • larger m or larger c → smaller temperature rise (Δ θ).
Specific heat capacity varies by materialWater has a considerably larger specific heat capacity than aluminium, iron or copper. The table supplies the approximate values and units.Specific heat capacity varies by materialMaterialSpecific heat capacity (J kg⁻¹ K⁻¹)
Approximate room-temperature values for comparison, not a required memorisation list. For equal masses receiving equal energy, without changing state, water has a smaller temperature rise than these metals.
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 mass and power actually heating the sample, it warms more slowly
  • for the same mass and energy-transfer rate out of the sample, its temperature falls more slowly

These comparisons assume no change of state and approximately constant c. Comparing different-sized objects requires their heat capacities, C = mc, rather than c alone.

Measuring specific heat capacity

The practical measurement lesson explains how to collect mass, heater-energy and temperature data, then evaluate unwanted energy transfers.

Using a temperature–time graph

Find a gradient from repeated readings in the temperature–time graph activity. Choose an interval with a steady response; the first readings may be affected by heater and probe lag.

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.

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:
    • heater energy E = Pt
    • set Q = E only if all that energy raises the sample’s temperature
  • State the ideal assumption: negligible energy transfer to the surroundings and negligible energy used to warm the heater, probe or container before setting 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

Worked 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.

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).

Show 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

Spot the mistake 3

Find the mass heated

About 5 min

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

Show 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

Finding specific heat capacity from heater power

Use the full worked example in Measuring specific heat capacity.

Try it yourself 4

Temperature rise from power

Minimal support

Changed-unknown transfer

A 300 W heater warms 0.50 kg of water initially at 20°C for 4.0 min. Assuming no unwanted energy transfer and using c = 4200 J kg⁻¹ K⁻¹, find the temperature rise. Assume the water does not change state.

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.

Show 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 and neither changes state. Treat each specific heat capacity as constant. Block A has a larger specific heat capacity than block B. The same amount of energy raises the temperature of each block. 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. Compare the energy released by the water alone, treating its specific heat capacity as constant and neglecting evaporation. Explain.

Show answer

The water releases the same amount of energy in each case. This does not compare energy released by the cups, whose heat capacities are not given.

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

Use Q = mcΔθ to solve a temperature-change problem, stating which body receives or releases the energy. For state changes at constant temperature, use specific latent heat. Continue practice in the Thermal Properties check.

Syllabus and review details