Required practical - Specific heat capacity Revision Notes for AQA GCSE Physics

Required practical - Specific heat capacity

What is specific heat capacity?

Specific heat capacity tells us how much energy we need to heat up 1 kg of a substance by 1°C. It's different for every material. Water has a high specific heat capacity, which means it takes lots of energy to heat it up.

infoNote

The word "specific" means "for 1 kg of mass" - this is important to remember when working with the formula!

Aim of the practical

We want to find out the specific heat capacity of water. We also look at what happens when ice melts.

Equipment you need

Water
Thermometer
Electric heater
Power supply
Insulation (to wrap around the beaker)
Beaker
Electronic balance

Method - step by step

Step 1: Set up your equipment as shown in the diagram. The thermometer goes in the water, and the electric heater heats the water from below.

Step 2: Use the electronic balance to measure the mass (m) of the water.

Step 3: Record the voltage (V) from the power supply and the current (I) through the heater.

Step 4: Take temperature readings every 30 seconds. Keep going until the water reaches your target temperature. Record the starting temperature and final temperature to work out the temperature change (Δθ).

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Worked Example: Taking Measurements

Measure mass of water: m = 0.5 kg
Set voltage: V = 12 V
Record current: I = 2 A
Initial temperature: 20°C
Final temperature: 30°C
Temperature change: Δθ = 10°C
Time taken: t = 180 seconds

Getting accurate results

chatImportant

Use insulation around your beaker. This stops heat energy escaping to the surroundings. Without insulation, your answer will be wrong because some energy is wasted.

Read the thermometer carefully. Make sure your eyes are level with the scale. Take readings regularly and check the thermometer is in the middle of the liquid.

Working out the results

Plot a graph of temperature against time. The graph should show the temperature rising steadily.

Use this equation to calculate specific heat capacity:

$c = \frac{V \times I \times t}{m \times \Delta\theta}$

Where:

$c$ = specific heat capacity (J/kg°C)
$V$ = voltage (V)
$I$ = current (A)
$t$ = time (s)
$m$ = mass of water (kg)
$\Delta\theta$ = temperature change (°C)

lightbulbExample

Worked Example: Calculating Specific Heat Capacity

Using the measurements from earlier:

V = 12 V, I = 2 A, t = 180 s
m = 0.5 kg, Δθ = 10°C

Step 1: Calculate energy supplied Energy = V × I × t = 12 × 2 × 180 = 4320 J

Step 2: Apply the formula $c = \frac{4320}{0.5 \times 10} = \frac{4320}{5} = 864 \text{ J/kg°C}$

What you should find

The specific heat capacity of water is about 4200 J/kg°C. This means you need 4200 joules of energy to heat 1 kg of water by 1°C.

Understanding ice melting

infoNote

When you heat ice, something interesting happens:

Ice warms up until it reaches 0°C
Then it melts, but the temperature stays at 0°C during melting
Once all the ice has melted, the water temperature starts rising again

This happens because energy goes into changing the state (solid to liquid) rather than increasing temperature.

Remember!

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Key Points to Remember:

Specific heat capacity = energy needed to heat 1 kg of a substance by 1°C
Water's specific heat capacity = 4200 J/kg°C
Use insulation to prevent heat loss and get accurate results
Take regular temperature readings every 30 seconds
The equation is $c = \frac{V \times I \times t}{m \times \Delta\theta}$

Required practical - Specific heat capacity (AQA GCSE Physics): Revision Notes