Heat Transfer (SQA National 5 Physics): Revision Note

100.0g of water is cooled in a fridge from 20.0 °C to 5.00 °C.

The specific heat capacity of water is 4180 J kg^-1 °C^-1.

Determine the amount of electrical energy used by the fridge to cool the water.

Answer:

Step 1: Convert the mass of water to SI units

$100\mathrm{g}\times \frac{1\mathrm{kg}}{1000\mathrm{g}}=0.100\mathrm{kg}$

Step 2: Determine the change in temperature

$∆T=20.0-5.00$

$∆T=15.0°\mathrm{C}$

Step 3: Apply the principle of conservation of energy

• The electrical energy used by the fridge to cool the water is equal to the magnitude of the heat energy transferred away from the water

$E_{fridge}=E_h=cm∆T$

Step 4: Substitute in the known values to calculate

$E_{fridge}=cm∆T$

$E_{fridge}=4180\times 0.100\times 15.0$

$E_{fridge}=6270\mathrm{J}$

Remember to always round your final answer to the correct number of significant figures, and to provide the unit. Students often lose marks in the exam for not doing so.

Remember that the calculated value can only be given to the same precision as the least precise value used in the calculation.

The least precise input value in this worked example is 3 s.f.; the answer is also given to 3 s.f.