(a) An electric water pump is powered by the 230 V mains supply - Edexcel - GCSE Physics - Question 5 - 2022 - Paper 1

Component X is likely a fuse. Its purpose is to protect the appliance and the wiring by breaking the circuit if the current exceeds a safe level, preventing overheating and potential fires.

To calculate the current, we will use the formula:

$I = \frac{E}{V \times t}$

Where:

• E = 9000 J
• V = 230 V
• t = 1 minute = 60 seconds

First, convert the time from minutes to seconds:

$t = 1 \times 60 = 60 \text{ seconds}$

Now substitute the values into the equation:

$I = \frac{9000 \text{ J}}{230 \text{ V} \times 60 \text{ s}}$

Calculating yields:

$I = \frac{9000}{13800} \approx 0.65 A$

The useful energy transferred to the water is less than the total energy supplied to the pump due to energy losses. Some energy is dissipated as thermal energy to the surroundings, due to friction and inefficiencies in the pump. In this case, the difference is 600 J, indicating energy was lost rather than used effectively.

To calculate the efficiency, we will use the equation:

$\text{efficiency} = \frac{\text{useful energy transferred by the pump}}{\text{total energy supplied to the pump}}$

Substituting the provided values:

$\text{efficiency} = \frac{8400 ext{ J}}{9000 ext{ J}}$

Calculating the efficiency:

$\text{efficiency} \approx 0.93 \text{ or } 93\%$