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.
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The core idea
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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) |
|---|---|
| water | 4200 |
| aluminium | 900 |
| iron | 450 |
| copper | 390 |
Data table
| Material | c |
|---|---|
| Water | 4200 |
| Aluminium | 900 |
| Iron | 450 |
| Copper | 390 |
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
- Measure the block’s mass m with a balance.
- Insert the heater and thermometer into their holes. Add a little thermal-contact material if instructed, so air gaps do not slow energy transfer.
- Wrap the block in insulation and record its initial temperature.
- Connect an ammeter in series with the heater and a voltmeter across it. Switch on and record I, V and heating time t.
- 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.
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
Problem
Study the worked solution
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⁻¹Scale 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⁻¹Find 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)
Problem
Use temperature change, not final temperature
Hints
Hint 1: find the rise
Hint 2: rearrange before substituting
View solution step by step
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 KCalculate 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
Learner response
Make mass the subject first
View solution step by step
Rearrange for mass
Method
Divide energy by specific heat capacity and temperature change.Reason
Mass multiplies both factors in Q = mcΔθ.Working
m = Q/cΔθSubstitute 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
Examination question
Show energy, temperature change, calculation and assumption
View solution step by step
Find supplied energy
1 markMethod
Multiply power by heating time.Reason
One watt is one joule per second.Working
Q = Pt = (50)(90) = 4500 JFind temperature change
1 markMethod
Subtract the initial reading.Reason
Q = mcΔθ uses temperature rise.Working
Δθ = 45-20 = 25 KCalculate and qualify the result
3 marksMethod
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
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark energy, temperature rise, substitution, result and assumption.
Challenge 5
Temperature rise from power
Changed-unknown transfer
Convert time before combining equations
Hints
Hint 1: match watts to seconds
Hint 2: combine energy equations
View solution step by step
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 JFind 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