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11.2Thermal Energy

Specific Heat Capacity and Thermal Capacity — American System Physics Revision Notes & Topical Questions

10 min read 7 parts Lesson 4 of 5 1 quiz

📚 American System Physics Hub / Topical Revision
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Lesson summary

Define specific heat capacity.Energy change = mass × specific heat capacity × temperature change: ΔE = mcΔθ (J, kg, J/(kg °C), °C).Rearrangedc = ΔE / (mΔθ); Δθ = ΔE / (mc); m = ΔE / (cΔθ).Thermal capacity of an objectC = mc (J/°C).Electrical energy suppliedΔE = VIt = Pt (V in vol…

  • Method of mixtures (no losses)
  • Mass must be in kg
  • Sources of error
  • Specific heat capacity (c)
  • Thermal (heat) capacity
  • Temperature change (Δθ or ΔT)

Section 1 of 7Specific Heat Capacity and Thermal Capacity

Essential knowledge
11.2 Specific Heat Capacity and Thermal C… concept map: Rearranged, Thermal capacity of an object, Electrical energy supplied, Method of mixtures (no losses), Mass must be in kg
Visual guide to the key ideas explained in this lesson.
Heating an object increases its internal energy. How much its temperature rises depends on the energy supplied, its mass and the material (its specific heat capacity).
A substance with a high specific heat capacity needs a lot of energy to warm up and releases a lot as it cools. Water's value (≈4200 J/(kg °C)) is unusually high, which is why it is used in car cooling systems, central heating radiators and hot-water bottles.
Coastal regions have milder climates than inland areas because the sea warms and cools slowly compared with land.
The thermal capacity of an object (mc) is the energy to raise the whole object by 1 °C. A large block has a larger thermal capacity than a small block of the same material.
In the electrical method, energy supplied is found from the heater's power multiplied by time, or from a voltmeter and ammeter (ΔE = VIt), or from a joulemeter.
Sources of error
Sources of error energy escapes to the surroundings, the heater and thermometer also absorb energy,
  • and the temperature may continue to rise after switching off.
  • Insulating the block, adding a lid, using a drop of oil to improve thermal contact with the thermometer,
  • and recording the maximum temperature all improve accuracy.
Because some energy is lost, the temperature rise measured is smaller than it should be, so the calculated c (= ΔE/mΔθ) is too large.
A graph of temperature against time for a constant-power heater is a straight line at first (ignoring losses); its gradient Δθ/t = P/(mc), so c = P ÷ (m × gradient).
In the method of mixtures, the energy lost by the hotter substance equals the energy gained by the cooler one (if no energy escapes), and both finish at the same final temperature.
At a change of state the temperature stops rising even though energy is still supplied, so ΔE = mcΔθ only applies while the substance stays in one state.

Section 2 of 7Process & sequence

1
Measure the mass m of an aluminium block (or of water in a beaker) with a balance.
2
Insert an immersion heater and a thermometer; wrap the block in insulation and record the starting temperature.
3
Connect the heater to a power supply with an ammeter and voltmeter (or a joulemeter); switch on and start a stopwatch.
4
After a measured time t (e.g. 10 minutes) switch off, and record the highest temperature reached.
5
Calculate energy supplied ΔE = VIt (or read from the joulemeter) and temperature rise Δθ.
6
Calculate c = ΔE ÷ (mΔθ); compare with the accepted value and explain why the result is usually too high (energy lost to surroundings).

Finding c from a heating graphwrite these steps in this order; each step is a marking point.

1
Heat the insulated block with a known constant power P.
2
Record temperature every 30 s and plot temperature (y) against time (x).
3
Draw a line of best fit through the straight section.
4
Measure the gradient Δθ/Δt in °C/s.
5
Calculate c = P ÷ (m × gradient).

Section 3 of 7Compare

Specific heat capacities of common materials
Substancec / J/(kg °C)Consequence
Water4200Heats and cools slowly; good coolant and store
Ice2100About half that of liquid water
Aluminium900Warms faster than water
Concrete≈ 880Used in storage heaters and thermal mass buildings
Iron/steel≈ 450Heats quickly (pans)
Copper385Heats very quickly
Lead130Tiny energy per degree

Values are approximate; always use those given in a question.

Section 4 of 7Worked examples

How much energy is needed to heat 2.5 kg of water from 20 °C to 80 °C? (c = 4200 J/(kg °C))
  • Δθ = 80 − 20 = 60 °C. ΔE = mcΔθ = 2.5 × 4200 × 60 = 630 000 J = 6.3 × 105 J (630 kJ).
    What the symbols mean: E, Q or q = energy transferred (J) · m = mass (kg; g in chemistry calorimetry) · c = specific heat capacity (J kg−1 °C−1; water 4.18 J g−1 °C−1) · ΔT or Δθ = temperature change (°C or K)
1000 J/(kg °C)
A 1.0 kg aluminium block is heated by a 50 W heater for 300 s. Its temperature rises from 21 °C to 36 °C. Calculate c for aluminium.
  • ΔE = Pt = 50 × 300 = 15 000 J.
    • Δθ = 36 − 21 = 15 °C. c = ΔE ÷ (mΔθ) = 15 000 ÷ (1.0 × 15) = 1000 J/(kg °C) (≈1.0 × 103).
    • This is higher than the accepted 900 J/(kg °C) because some energy was lost to the surroundings.
48 °C
0.20 kg of water at 90 °C is mixed with 0.30 kg of water at 20 °C. Find the final temperature, ignoring losses.
  • Energy lost by hot = energy gained by cold: 0.20 × 4200 × (90 − T) = 0.30 × 4200 × (T − 20).
    • 0.20(90 − T) = 0.30(T − 20) → 18 − 0.2T = 0.3T − 6 → 24 = 0.5T → T = 48 °C.

Section 5 of 7Key terms

Section 6 of 7Command terms

Command terms
Identify / Describe (FRQ)
A short, specific statement; describing requires relevant characteristics, not a single word.
Explain / Justify (FRQ)
Give the cause-and-effect reasoning or evidence; a restated fact does not score.
Calculate
Show setup, substitution and the answer with units.
Predict
State the expected outcome; add a reason only if asked to justify.
MCQ technique
Read every option; eliminate distractors that are true statements but do not answer the question; watch for NOT/EXCEPT.

Section 7 of 7Exam technique

Keywords the mark scheme looks for in this lesson

Keywords the mark scheme looks for in this lesson
Specific heat capacity (c) · Thermal (heat) capacity · Temperature change (Δθ or ΔT) · Energy supplied (ΔE) · Joulemeter · Immersion heater · Insulation (lagging) · Coolant · Heat sink · Method of mixtures · Power of heater (P) · Systematic error from heat loss
Learn each definition in the Key words tab exactly; a missing keyword in a definition usually loses the mark.

How to get full marks (examiner guidance)

1
Answer exactly what is asked; extra correct material does not earn credit, and contradictory statements can cancel credit (AP FRQ).
2
Use precise vocabulary, never “it/stuff/things”.
3
Show units and working in calculations.
4
In MCQs, predict the answer before reading the options, then eliminate.

Words to avoid

1
specific heat capacity is the heat needed to raise the temperatureit is the energy required per unit mass per unit temperature increase (per kg per °C).
2
Don't use mass in grams in ΔE = mcΔθ — Write: convert to kilograms first.
3
heat losses make c smallerheat losses make the calculated c larger than the true value.
4
water heats up slowly because it is a poor conductorwater has a high specific heat capacity, so it needs more energy per kg per °C.
5
the unit of c is J/°CJ/(kg °C); J/°C is the unit of thermal capacity.

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