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How much energy does it take to heat a material, and how do we measure it?

Specific heat capacity: the energy needed to raise the temperature of a substance, the equation linking change in thermal energy to mass, specific heat capacity and temperature change, and the required practical.

A focused answer to AQA GCSE Physics 4.1.1, covering the meaning of specific heat capacity, the equation linking thermal energy, mass, specific heat capacity and temperature change, and the required practical to measure it.

Generated by Claude Opus 4.88 min answer

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  1. What this dot point is asking
  2. What specific heat capacity means
  3. The equation
  4. Required practical: measuring specific heat capacity
  5. Try this

What this dot point is asking

AQA wants you to define specific heat capacity, use the equation linking change in thermal energy to mass, specific heat capacity and temperature change, and describe the required practical for measuring the specific heat capacity of a material.

What specific heat capacity means

A material with a high specific heat capacity, such as water (c4200J/kgCc \approx 4200\,J/kg\,^{\circ}C), needs a large amount of energy to warm up and stores a large amount of energy. This is why water is used in central-heating systems and as a coolant. The same property explains why coastal areas have milder climates than places far inland: the sea warms up and cools down slowly because of water's high specific heat capacity, so it moderates the temperature of nearby land. By contrast, metals such as copper and aluminium have low specific heat capacities (around 385385 and 900J/kgC900\,J/kg\,^{\circ}C), so they heat up quickly for a given energy input, which is useful in cooking pans where you want a fast temperature response.

It is worth being clear about what specific heat capacity is not. It is not the same as the total heat capacity of an object, which depends on how much of the material there is. A large swimming pool and a cup both contain water with the same specific heat capacity, but the pool needs far more energy to warm by one degree because it has much more mass. The specific heat capacity is a property of the material itself, independent of the amount.

The equation

Required practical: measuring specific heat capacity

You heat a known mass of the material (a metal block or a beaker of water) with an electric immersion heater. A joulemeter (or voltmeter and ammeter with a timer) measures the energy supplied, and a thermometer records the temperature rise.

In the standard version of the practical, a cylindrical metal block has two holes drilled in it, one for the heater and one for the thermometer (a little oil or water in the thermometer hole improves thermal contact). You record the temperature, switch on the heater for a measured time while recording the energy supplied, and record the final temperature. The energy supplied electrically can be found from E=PtE = Pt if you know the power, or read directly from a joulemeter. Plotting temperature against energy supplied gives a straight line whose gradient can be used to find the specific heat capacity, which often gives a more reliable result than a single pair of readings.

Try this

Q1. Define specific heat capacity. [2 marks]

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

Q2. Calculate the energy needed to raise the temperature of a 2kg2\,kg aluminium block (c=900J/kgCc = 900\,J/kg\,^{\circ}C) by 15C15\,^{\circ}C. [3 marks]

  • Cue. ΔE=mcΔθ=2×900×15=27000J\Delta E = mc\Delta\theta = 2 \times 900 \times 15 = 27000\,J.

Exam-style practice questions

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

AQA 20195 marksA 2.0kg2.0\,\text{kg} block of copper is heated from 18C18\,^{\circ}\text{C} to 58C58\,^{\circ}\text{C}. The specific heat capacity of copper is 385J/kgC385\,\text{J/kg}\,^{\circ}\text{C}. Calculate the energy needed, then explain why the energy actually supplied by the heater in an experiment would be greater than this value.
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Use ΔE=mcΔθ\Delta E = mc\Delta\theta. The temperature change is Δθ=5818=40C\Delta\theta = 58 - 18 = 40\,^{\circ}\text{C} (1 mark). So ΔE=2.0×385×40=30,800J\Delta E = 2.0 \times 385 \times 40 = 30{,}800\,\text{J}, which is about 3.08×104J3.08 \times 10^{4}\,\text{J} or 30.8kJ30.8\,\text{kJ} (2 marks). The energy supplied by the heater would be greater because some energy is transferred to the surroundings by heating (the block and air around it are not perfectly insulated), so not all the electrical energy supplied goes into the copper's thermal store (2 marks). Markers reward the temperature difference, the correct substitution and value, and the explanation referring to energy lost to the surroundings. Insulating the block reduces this loss.

AQA 20213 marksDefine specific heat capacity, and explain why water, which has a high specific heat capacity, is used as the coolant in many heating and cooling systems.
Show worked answer →

The specific heat capacity of a substance is the energy needed to raise the temperature of 1kg1\,\text{kg} of it by 1C1\,^{\circ}\text{C} (1 mark). Water has a high specific heat capacity, so it can absorb (or release) a large amount of energy for only a small change in temperature (1 mark). This makes it a good coolant because it can carry a lot of energy away from a hot part without its own temperature rising too much, and a good store for central heating because it holds a lot of energy to release as it cools (1 mark). Markers reward the per-kilogram, per-degree definition and the link between high specific heat capacity and the ability to transfer large amounts of energy with small temperature changes.

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