Energy changes in systems Revision Notes for AQA GCSE Physics Combined Science

Energy changes in systems

When objects heat up or cool down, energy moves into or out of the system. You can work out exactly how much energy is involved by using something called specific heat capacity.

What is specific heat capacity?

Specific heat capacity is the amount of thermal energy needed to raise the temperature of 1 kg of a material by 1°C.

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Different materials need different amounts of energy to heat up:

Water needs lots of energy to heat up (that's why kettles take time to boil)
Metals need less energy to heat up (that's why metal spoons get hot quickly in tea)

The specific heat capacity equation

You can calculate energy changes using this formula:

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The Specific Heat Capacity Formula:

$\Delta E = m \times c \times \Delta \theta$

Where:

$\Delta E$ = change in thermal energy (in joules, J)
$m$ = mass (in kg)
$c$ = specific heat capacity (in J/kg°C)
$\Delta \theta$ = change in temperature (in °C)

Typical specific heat capacities

Here are some common values you should know:

Material	Specific heat capacity (J/kg°C)
Air	1005
Water	4200
Aluminium	900
Concrete	880

Key fact: Water has a very high specific heat capacity at 4200 J/kg°C. This means 1 kg of water needs 4200 J of energy to increase its temperature by just 1°C.

Finding specific heat capacity in the lab

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Laboratory Apparatus Setup:

You can measure specific heat capacity using this apparatus:

A metal block with a heater inside it
A thermometer to measure temperature
An ammeter to measure current
A low voltage power supply
Insulating material (lagging) around the block to reduce heat loss

The insulation stops energy being wasted to the surroundings, making your results more accurate.

Worked example 1: Heating metal

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Worked Example: Calculating Thermal Energy Required

Question: A 0.8 kg metal block has a specific heat capacity of 900 J/kg°C. It's heated from 20°C to 80°C. How much thermal energy is needed?

Solution:

$m = 0.8$ kg
$c = 900$ J/kg°C
$\Delta \theta = 80°C - 20°C = 60°C$

Using the formula: $\Delta E = m \times c \times \Delta \theta$

$\Delta E = 0.8 \times 900 \times 60 = 43,200$ J

Answer: 43,200 J of thermal energy is needed

Worked example 2: Finding specific heat capacity

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Worked Example: Determining Specific Heat Capacity

Question: A 2.8 kg metal block is heated from 18°C to 52°C using 300,000 J of electrical energy. What is the specific heat capacity?

Solution:

$m = 2.8$ kg
$\Delta E = 300,000$ J
$\Delta \theta = 52°C - 18°C = 34°C$

Rearranging the formula: $c = \frac{\Delta E}{m \times \Delta \theta}$

$c = \frac{300,000}{2.8 \times 34}$

$c = \frac{300,000}{95.2} = 3151$ J/kg°C

Answer: The specific heat capacity is 3151 J/kg°C

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Key Points to Remember:

Specific heat capacity is the energy needed to heat 1 kg of material by 1°C
Water has the highest common specific heat capacity at 4200 J/kg°C
Use the formula $\Delta E = mc\Delta \theta$ to calculate energy changes
Always include units in your answers (J for energy, kg for mass, °C for temperature)
Insulation reduces heat loss in experiments, making results more accurate

Energy changes in systems (AQA GCSE Physics Combined Science): Revision Notes