Specific Heat Capacity

This lesson covers specific heat capacity (SHC) as required by the Edexcel GCSE Physics specification (1PH0), Topic 3: Conservation of Energy. You need to understand what SHC means, use the equation in calculations, and know the core practical for investigating SHC.

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What Is Specific Heat Capacity?

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

Every substance has its own SHC — it is a property of the material, not the object.

The Equation

$\Delta E = m c \Delta \theta$ΔE=mcΔθ

Where:

• ΔE = change in thermal energy (joules, J)
• m = mass (kilograms, kg)
• c = specific heat capacity (joules per kilogram per degree Celsius, J/kg°C)
• Δθ = change in temperature (degrees Celsius, °C)

Units

Quantity	Unit
Energy (ΔE)	joules (J)
Mass (m)	kilograms (kg)
Specific heat capacity (c)	J/kg°C (or J/kg K)
Temperature change (Δθ)	°C (or K)

Exam Tip: Note that Δθ (delta theta) means change in temperature — not the final temperature. Δθ = final temperature − initial temperature. A very common mistake is to use the final temperature alone.

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SHC Values for Common Materials

Material	SHC (J/kg°C)	What This Means
Water	4200	It takes 4200 J to heat 1 kg of water by 1 °C
Aluminium	900	It takes 900 J to heat 1 kg of aluminium by 1 °C
Copper	390	It takes 390 J to heat 1 kg of copper by 1 °C
Iron	450	It takes 450 J to heat 1 kg of iron by 1 °C
Lead	130	It takes 130 J to heat 1 kg of lead by 1 °C
Oil	2100	It takes 2100 J to heat 1 kg of oil by 1 °C

What Does a High SHC Mean?

• A high SHC means the substance needs a lot of energy to change its temperature — it heats up and cools down slowly.
• A low SHC means the substance needs less energy to change its temperature — it heats up and cools down quickly.

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Why Water Has a High SHC

Water has an unusually high SHC (4200 J/kg°C). This has important real-world implications:

• Central heating systems use water because it can store and transfer a large amount of thermal energy.
• Coastal climates are milder than inland climates because the sea absorbs and releases large amounts of energy without large temperature changes.
• Car cooling systems use water because it can absorb a lot of waste thermal energy from the engine.
• Water is used as a coolant in power stations and industrial processes.

Exam Tip: If asked "Why is water used as a coolant?", state that water has a high specific heat capacity, so it can absorb a large amount of energy for a small temperature rise, making it effective at removing waste heat.

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Worked Examples

Example 1: Calculating Energy

Question: How much energy is needed to heat 2 kg of water from 20 °C to 80 °C? (c = 4200 J/kg°C)

Solution:

$\Delta \theta = 80 - 20 = 60 \text{ °C}$Δθ=80−20=60 °C $\Delta E = m c \Delta \theta = 2 \times 4200 \times 60 = 504{,}000 \text{ J} = 504 \text{ kJ}$ΔE=mcΔθ=2×4200×60=504,000 J=504 kJ

Example 2: Finding Temperature Change

Question: 18 000 J of energy is supplied to 0.5 kg of aluminium. What is the temperature rise? (c = 900 J/kg°C)

Solution:

$\Delta \theta = \frac{\Delta E}{m c} = \frac{18{,}000}{0.5 \times 900} = \frac{18{,}000}{450} = 40 \text{ °C}$Δθ=mcΔE​=0.5×90018,000​=45018,000​=40 °C

Example 3: Finding Mass

Question: A copper block is heated by 50 °C using 9750 J of energy. What is the mass of the block? (c = 390 J/kg°C)

Solution:

$m = \frac{\Delta E}{c \Delta \theta} = \frac{9750}{390 \times 50} = \frac{9750}{19{,}500} = 0.5 \text{ kg}$m=cΔθΔE​=390×509750​=19,5009750​=0.5 kg

Example 4: Finding SHC

Question: 13 500 J of energy heats 3 kg of a substance by 10 °C. What is its specific heat capacity?

Solution:

$c = \frac{\Delta E}{m \Delta \theta} = \frac{13{,}500}{3 \times 10} = \frac{13{,}500}{30} = 450 \text{ J/kg°C}$c=mΔθΔE​=3×1013,500​=3013,500​=450 J/kg°C

This suggests the substance is iron (c = 450 J/kg°C).

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Core Practical: Investigating Specific Heat Capacity

Aim

To measure the specific heat capacity of a material (typically a metal block — aluminium or copper).

Equipment

• Metal block with two holes (one for the heater, one for the thermometer)
• Immersion heater (electric heater)
• Thermometer (or temperature sensor)
• Joulemeter (or ammeter, voltmeter, and stopwatch)
• Power supply
• Insulation (e.g., cotton wool or bubble wrap around the block)
• Balance (to measure the mass of the block)

Method

1. Measure the mass of the metal block using a balance.
2. Insert the immersion heater into one hole and the thermometer into the other.
3. Wrap the block in insulation to reduce heat loss to the surroundings.
4. Record the starting temperature (θ₁).
5. Switch on the heater and the joulemeter. Heat the block for a set time (e.g., 10 minutes).
6. Record the energy supplied (from the joulemeter) and the final temperature (θ₂).
7. Calculate: $c = \frac{\Delta E}{m \Delta \theta}$c=mΔθΔE​

Alternative Method (Without a Joulemeter)

If no joulemeter is available, use an ammeter and voltmeter:

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

Where V = voltage (V), I = current (A), t = time (s).

Variables

Variable	Type	Detail
Energy supplied	Independent (controlled via time)	Measured with joulemeter or V × I × t
Temperature change	Dependent	Measured with thermometer
Mass of block, type of block	Control	Same block throughout

Sources of Error