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Why do different substances need different amounts of energy to change temperature or state, and how do we calculate that energy?

Define and apply specific heat capacity and specific latent heat to calculate energy transfers during temperature changes and changes of state

A focused answer to the H2 Physics learning outcome on specific heat capacity and specific latent heat. The defining equations, why latent heat involves no temperature change, and multi-stage heating and phase-change calculations.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to define and apply specific heat capacity (energy to change temperature) and specific latent heat (energy to change state), and to handle problems that combine heating and phase change. The key conceptual point is that a change of state happens at constant temperature, because the energy goes into breaking molecular bonds rather than speeding molecules up.

The answer

Specific heat capacity

The specific heat capacity cc of a substance is the energy required to raise the temperature of 1 kg1\ \text{kg} of it by 1 K1\ \text{K}:

Q=mcΔθQ = mc\,\Delta\theta

where mm is the mass, cc the specific heat capacity (J kg1K1\text{J kg}^{-1}\text{K}^{-1}) and Δθ\Delta\theta the temperature change. Water has an unusually high specific heat capacity (4200 J kg1K14200\ \text{J kg}^{-1}\text{K}^{-1}), which is why it is an effective coolant and moderates climate.

Specific latent heat

The specific latent heat LL of a substance is the energy required to change the state of 1 kg1\ \text{kg} of it without any change in temperature:

Q=mLQ = mL

  • The specific latent heat of fusion applies to melting and freezing.
  • The specific latent heat of vaporisation applies to boiling and condensing.

The latent heat of vaporisation is generally much larger than that of fusion, because separating molecules completely (liquid to gas) requires more energy than loosening them (solid to liquid).

Why temperature is constant during a change of state

During melting or boiling, the supplied energy goes into increasing the molecular potential energy by breaking or weakening intermolecular bonds, not into increasing molecular kinetic energy. Since temperature measures average molecular kinetic energy, the temperature stays constant until the change of state is complete.

Multi-stage problems

Heating a substance through a phase change requires adding the energies for each stage in turn: temperature change (mcΔθmc\Delta\theta), then change of state (mLmL), then any further temperature change. A heating curve (temperature against energy supplied) shows sloping sections (temperature rising) separated by flat plateaus (state changing at constant temperature).

Examples in context

Example 1. Why sweating cools you. As sweat evaporates from your skin, it absorbs its latent heat of vaporisation from your body. Because that latent heat is large, even a small mass of evaporating sweat removes a significant amount of energy, cooling you efficiently. This is the same physics that makes steam burns more severe than hot-water burns.

Example 2. Climate moderation by the sea. Water's high specific heat capacity means coastal regions warm and cool more slowly than inland regions. The sea absorbs and releases large amounts of energy for modest temperature changes, smoothing out daily and seasonal extremes near the coast.

Try this

Q1. Define specific heat capacity and state its SI unit. [2 marks]

  • Cue. Energy to raise the temperature of 1 kg1\ \text{kg} by 1 K1\ \text{K}; unit J kg1K1\text{J kg}^{-1}\text{K}^{-1}.

Q2. Find the energy to heat 0.40 kg0.40\ \text{kg} of aluminium (c=900 J kg1K1c = 900\ \text{J kg}^{-1}\text{K}^{-1}) from 25 C25\ ^\circ\text{C} to 75 C75\ ^\circ\text{C}. [2 marks]

  • Cue. Q=mcΔθ=0.40×900×50=1.8×104 JQ = mc\Delta\theta = 0.40 \times 900 \times 50 = 1.8 \times 10^4\ \text{J}.

Q3. Explain, in terms of molecules, why the temperature of melting ice stays at 0 C0\ ^\circ\text{C} while energy is supplied. [3 marks]

  • Cue. The supplied energy increases molecular potential energy by breaking the bonds of the solid lattice, not the kinetic energy; since temperature measures average kinetic energy, it stays constant until melting is complete.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksAn electric heater of power 50 W50\ \text{W} heats 0.30 kg0.30\ \text{kg} of water. The specific heat capacity of water is 4200 J kg1K14200\ \text{J kg}^{-1}\text{K}^{-1}. (a) Find the energy needed to raise the temperature from 20 C20\ ^\circ\text{C} to 80 C80\ ^\circ\text{C}. (b) Find the time taken, assuming no heat loss. (c) Explain how heat loss would affect the actual time.
Show worked answer →

(a) Energy: Q=mcΔθ=0.30×4200×(8020)=0.30×4200×60=7.56×104 JQ = mc\Delta\theta = 0.30 \times 4200 \times (80 - 20) = 0.30 \times 4200 \times 60 = 7.56 \times 10^4\ \text{J}.

(b) Time: t=QP=7.56×10450=1512 s1.5×103 st = \dfrac{Q}{P} = \dfrac{7.56 \times 10^4}{50} = 1512\ \text{s} \approx 1.5 \times 10^3\ \text{s} (about 2525 minutes).

(c) With heat loss to the surroundings, some of the heater's energy is wasted, so more than the calculated energy must be supplied. The actual heating time would be longer than 1512 s1512\ \text{s}.

Markers reward Q=mcΔθQ = mc\Delta\theta with the correct temperature change, time as energy over power, and the recognition that heat loss lengthens the actual time.

Original5 marksCalculate the total energy required to convert 0.20 kg0.20\ \text{kg} of ice at 0 C0\ ^\circ\text{C} entirely into water at 40 C40\ ^\circ\text{C}. Specific latent heat of fusion of ice =3.3×105 J kg1= 3.3 \times 10^5\ \text{J kg}^{-1}; specific heat capacity of water =4200 J kg1K1= 4200\ \text{J kg}^{-1}\text{K}^{-1}.
Show worked answer →

The process has two stages.

Stage 1, melting the ice at 0 C0\ ^\circ\text{C} (no temperature change): Q1=mL=0.20×3.3×105=6.6×104 JQ_1 = mL = 0.20 \times 3.3 \times 10^5 = 6.6 \times 10^4\ \text{J}.

Stage 2, heating the resulting water from 0 C0\ ^\circ\text{C} to 40 C40\ ^\circ\text{C}: Q2=mcΔθ=0.20×4200×40=3.36×104 JQ_2 = mc\Delta\theta = 0.20 \times 4200 \times 40 = 3.36 \times 10^4\ \text{J}.

Total: Q=Q1+Q2=6.6×104+3.36×104=9.96×104 J1.0×105 JQ = Q_1 + Q_2 = 6.6 \times 10^4 + 3.36 \times 10^4 = 9.96 \times 10^4\ \text{J} \approx 1.0 \times 10^5\ \text{J}.

Markers reward splitting into melting (latent heat, constant temperature) and heating (specific heat capacity) stages, the correct formula in each, and the total.

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