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Why do different materials need different amounts of energy to heat up, and how do we calculate that energy?

Specific heat capacity: the energy needed to change the temperature of a material, the relationship linking energy, mass, specific heat capacity and temperature change, and using it in heating and mixing problems.

An SQA National 5 Physics answer on specific heat capacity, covering the meaning of specific heat capacity, the relationship linking energy, mass, specific heat capacity and temperature change, why water needs so much energy to heat, and how to use the relationship in heating and energy-transfer problems.

Generated by Claude Opus 4.810 min answer

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

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  1. What this key area is asking
  2. What specific heat capacity means
  3. The specific heat capacity relationship
  4. Why water's high value matters
  5. Measuring specific heat capacity
  6. Try this

What this key area is asking

The SQA wants you to explain what specific heat capacity means, use the relationship E=mcΔTE = mc\Delta T to find the energy needed to change a material's temperature, and apply it in heating and energy-transfer problems including experiments where energy is supplied by a heater.

What specific heat capacity means

Different materials have very different values. Water is unusually high at 4180 J kg−1 ∘C−14180 \text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1}, while metals are much lower, for example aluminium at about 902902 and copper at about 386 J kg−1 ∘C−1386 \text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1}. This is why a metal spoon heats up quickly in a hot drink but the water itself takes much longer.

The specific heat capacity relationship

The temperature change ΔT\Delta T is the final temperature minus the starting temperature. You can rearrange the relationship to find any quantity: c=EmΔTc = \frac{E}{m\Delta T} to find a specific heat capacity from an experiment, or ΔT=Emc\Delta T = \frac{E}{mc} to find a temperature rise.

Why water's high value matters

Because water has such a high specific heat capacity, it stores a great deal of heat without its temperature rising much. This is why water is used in central-heating systems and car-engine coolant, why the sea warms and cools more slowly than the land (giving coastal places milder weather), and why a hot-water bottle stays warm for hours. The same property means a lot of energy is needed to heat a kettle or a bath.

Measuring specific heat capacity

In a typical experiment, a known mass of a material is heated by an electrical heater of known power for a measured time, and the temperature rise is recorded. The energy supplied is found from E=PtE = Pt and put into E=mcΔTE = mc\Delta T, rearranged to c=EmΔTc = \frac{E}{m\Delta T}. Experimental values usually come out a little high because some of the heater's energy is lost to the surroundings rather than going into the material; lagging (insulation) reduces this loss.

Try this

Q1. State what is meant by the specific heat capacity of a material. [1 mark]

  • Cue. The energy needed to raise the temperature of 1 kg1 \text{ kg} of it by 1 ∘C1\,{}^{\circ}\text{C}.

Q2. Calculate the energy needed to heat 0.50 kg0.50 \text{ kg} of water by 20 ∘C20\,{}^{\circ}\text{C} (c=4180 J kg−1 ∘C−1c = 4180 \text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1}). [2 marks]

  • Cue. E=mcΔT=0.50×4180×20=41 800 JE = mc\Delta T = 0.50 \times 4180 \times 20 = 41\,800 \text{ J}.

Q3. State why coastal areas have milder weather than inland areas. [1 mark]

  • Cue. Water's high specific heat capacity means the sea warms and cools slowly.

Exam-style practice questions

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

SQA N5 style4 marksCalculate the energy needed to raise the temperature of 2.0 kg of water by 30 degrees Celsius. The specific heat capacity of water is 4180 J per kg per degree Celsius.
Show worked answer →

Use the relationship linking energy, mass, specific heat capacity and temperature change.

Relationship: E=mcΔTE = mc\Delta T.

Substitution: E=2.0×4180×30=250 800 JE = 2.0 \times 4180 \times 30 = 250\,800 \text{ J}.

Markers reward selecting E=mcΔTE = mc\Delta T, correct substitution of the mass, specific heat capacity and temperature change, and a final answer in joules (J\text{J}). The answer rounds to 2.5×105 J2.5 \times 10^5 \text{ J}.

SQA N5 style4 marksA 0.50 kg block of aluminium is heated by a 60 W heater for 2 minutes and its temperature rises by 30 degrees Celsius. Calculate the specific heat capacity of the aluminium found in this experiment.
Show worked answer →

First find the energy supplied: E=Pt=60×120=7200 JE = Pt = 60 \times 120 = 7200 \text{ J} (time converted to 120 s120 \text{ s}).

Then rearrange the heat relationship for the specific heat capacity: c=EmΔTc = \dfrac{E}{m\Delta T}.

Substitution: c=72000.50×30=720015=480 J kg−1 ∘C−1c = \dfrac{7200}{0.50 \times 30} = \dfrac{7200}{15} = 480 \text{ J kg}^{-1}\,{}^{\circ}\text{C}^{-1}.

Markers reward finding the energy from E=PtE = Pt, rearranging E=mcΔTE = mc\Delta T for cc, correct substitution, and the unit. The experimental value is a little high because some heat is lost to the surroundings.

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