The surface temperature of a swimming pool on a... - JAMB Physics 2023 Question

The rate at which energy is transferred by conduction from the surface to the bottom of the swimming pool (\(\frac{q}{t}\)) can be calculated using the formula:

\[
\frac{q}{t} = K \cdot A \cdot \frac{\Delta \theta}{L}
\]

where:
- \(K\) is the thermal conductivity of water (\(0.6071 \, \text{W/m} \cdot \text{K}\)),
- \(A\) is the surface area of the pool (\(620 \, \text{m}^2\)),
- \(\Delta \theta\) is the temperature difference (\(25^\circ \text{C} - 15^\circ \text{C} = 10^\circ \text{C}\)),
- \(L\) is the depth of the pool (\(1.5 \, \text{m}\)).

Substitute the values:

\[
\frac{q}{t} = 0.6071 \times 620 \times \frac{10}{1.5} = 2509.35 \, \text{W}
\]

Convert the result to kilowatts (\(\text{kW}\)):

\[
\frac{2509.35 \, \text{W}}{1000} = 2.50935 \, \text{kW}
\]

Approximately, the rate of energy transfer is \(2.5 \, \text{kW}\), so the correct option is: \(2.5 \, \text{kW}\).