Specific Heat Capacity

Q = mcΔθ; required practical; comparing materials

# Specific Heat Capacity — GCSE Physics

Why does a metal saucepan heat up quickly while water takes ages to boil? The answer lies in specific heat capacity — a property that tells us how much energy is needed to change the temperature of a material.


1. What Is Specific Heat Capacity?

Specific heat capacity (cc) is the amount of energy needed to raise the temperature of 1 kg of a substance by 1°C (or 1 K).

ΔE=m×c×Δθ\boxed{\Delta E = m \times c \times \Delta\theta}

Where:

  • ΔE\Delta E = energy transferred (J)
  • mm = mass (kg)
  • cc = specific heat capacity (J/kg°C)
  • Δθ\Delta\theta = change in temperature (°C)

Units of cc: J/kg°C (or J/kg·K — the values are the same since a 1°C change equals a 1 K change).

Common Values

Material Specific Heat Capacity (J/kg°C)
Water 4200
Ice 2100
Aluminium 900
Copper 390
Iron/Steel 450
Lead 130
Oil 2000
Concrete 880
Air 1000

Water has an exceptionally high specific heat capacity — this is why:

  • It takes a long time to boil water
  • Oceans moderate coastal temperatures
  • Water is used as a coolant in engines and power stations
  • Central heating systems use water to carry thermal energy

2. Rearranging the Equation

m=ΔEc×Δθc=ΔEm×ΔθΔθ=ΔEm×cm = \frac{\Delta E}{c \times \Delta\theta} \qquad c = \frac{\Delta E}{m \times \Delta\theta} \qquad \Delta\theta = \frac{\Delta E}{m \times c}


3. Required Practical: Investigating Specific Heat Capacity

Aim

To measure the specific heat capacity of a solid (e.g., aluminium block) or liquid (e.g., water).

Equipment

  • Insulated metal block (with holes for heater and thermometer) or beaker of water
  • Electric immersion heater
  • Thermometer (or temperature sensor)
  • Joulemeter (or ammeter + voltmeter + stopwatch)
  • Balance
  • Insulation (lagging)

Method (Metal Block)

  1. Measure the mass of the metal block
  2. Insert the heater and thermometer into the block
  3. Record the initial temperature
  4. Switch on the heater and record energy supplied (using joulemeter) and temperature at regular intervals
  5. After heating for a suitable time, switch off and record the highest temperature
  6. Calculate cc using the equation

Method (Water)

  1. Measure the mass of water in an insulated beaker
  2. Record initial temperature
  3. Heat with immersion heater, recording energy and temperature
  4. Calculate cc

Calculating Energy Without a Joulemeter

E=P×t=I×V×tE = P \times t = I \times V \times t

Where PP = power, II = current, VV = voltage, tt = time.

Sources of Error

  • Energy loss to surroundings → measured cc will be too high (more energy needed than expected)
  • Insulation reduces this error
  • Thermometer position — may not measure the hottest/average temperature
  • Thermal lag — temperature continues rising after heater is switched off

Worked Example: Example 1

Problem

Question: Calculate the energy needed to heat 2 kg of water from 20°C to 100°C. (c=4200c = 4200 J/kg°C)

ΔE=mcΔθ=2×4200×(10020)=2×4200×80=672,000 J=672 kJ\Delta E = mc\Delta\theta = 2 \times 4200 \times (100 - 20) = 2 \times 4200 \times 80 = 672{,}000 \text{ J} = 672 \text{ kJ}

Solution

Worked Example: Example 2

Problem

Question: 50,000 J of energy is supplied to a 5 kg aluminium block. Calculate the temperature rise. (c=900c = 900 J/kg°C)

Δθ=ΔEmc=50,0005×900=50,0004500=11.1°C\Delta\theta = \frac{\Delta E}{mc} = \frac{50{,}000}{5 \times 900} = \frac{50{,}000}{4500} = 11.1°\text{C}

Solution

Worked Example: Example 3

Problem

Question: A 0.3 kg copper block is heated from 25°C to 75°C. 5850 J of energy was transferred. Calculate the specific heat capacity.

c=ΔEmΔθ=58500.3×50=585015=390 J/kg°Cc = \frac{\Delta E}{m\Delta\theta} = \frac{5850}{0.3 \times 50} = \frac{5850}{15} = 390 \text{ J/kg°C}

Solution

Worked Example: Example 4

Problem

Question: A 2 kW heater heats 1.5 kg of water for 3 minutes. Calculate the temperature rise. (c=4200c = 4200 J/kg°C)

E=Pt=2000×180=360,000 JE = Pt = 2000 \times 180 = 360{,}000 \text{ J} Δθ=360,0001.5×4200=360,0006300=57.1°C\Delta\theta = \frac{360{,}000}{1.5 \times 4200} = \frac{360{,}000}{6300} = 57.1°\text{C}


Solution

5. Practice Questions

    1. Calculate the energy needed to heat 0.5 kg of copper from 20°C to 120°C. (c=390c = 390 J/kg°C) (2 marks)
    1. 10,000 J of energy is supplied to 2 kg of oil (c=2000c = 2000 J/kg°C). Calculate the temperature rise. (2 marks)
    1. Explain why water is used in car radiators as a coolant. (2 marks)
    1. A 3 kW heater is used to heat 5 kg of water from 15°C. How long will it take to reach 100°C? (3 marks)
    1. In a practical, a student supplies 8000 J to a 1 kg metal block and records a temperature rise of 20°C. Calculate the SHC and identify the metal. (3 marks)

    Answers

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Frequently Asked Questions

What does "specific" mean in specific heat capacity?

"Specific" means "per unit mass" — it's the heat capacity per kilogram. The total heat capacity of an object depends on its mass AND its specific heat capacity.

Is the answer always too high in the practical?

Usually yes, because energy escapes to the surroundings. You supply more energy than actually goes into the block/water, so the calculated cc is higher than the true value.

Why does water have such a high SHC?

Water molecules form strong hydrogen bonds. A lot of energy is needed to increase the kinetic energy of water molecules because some energy goes into breaking and reforming these intermolecular bonds.


Summary

  • Specific heat capacity = energy to raise 1 kg by 1°C
  • ΔE=mcΔθ\Delta E = mc\Delta\theta
  • Water has a very high SHC (4200 J/kg°C)
  • Required practical: heat a block/water, measure energy and temperature change
  • Key error: energy loss to surroundings (use insulation)

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