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

The purpose of component X is to ensure safety by preventing electric shock. It serves to ground any stray electricity, directing it to the earth instead of through the user, thereby minimizing the risk of harm.

To find the current (I), we can use the formula:

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

Where:

• E = 9000 J
• V = 230 V
• t = 1 minute = 60 seconds Substituting the values: $I = \frac{9000}{230 \times 60}$

Calculating: $I = \frac{9000}{13800} = 0.65217 A$

Thus, rounding, the current is approximately 0.65 A.

The useful energy transferred to the water is less than the total energy supplied because some of the energy is dissipated as thermal energy due to inefficiencies in the system. This loss can occur through heat produced in the pump mechanics and other resistive losses, leading to a difference of 600 J between the total energy supplied and the useful energy transferred.

The efficiency can be calculated using the formula:

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

Substituting the values: $\text{efficiency} = \frac{8400}{9000}$

Calculating the efficiency gives: $\text{efficiency} = 0.9333$

Thus, the efficiency is approximately 0.93 or 93%.