The surface temperature of a swimming pool on a warm day is 25ºC and the temperature at the bottom is 15ºC. If…

Analysis

To solve this problem, we need to calculate the rate of heat transfer (heat flow) through a liquid medium (water) due to conduction. We are given the temperature difference, the surface area, the thickness (depth), and the thermal conductivity of the material.

Concept

The rate of heat transfer by conduction is governed by Fourier’s Law of Heat Conduction, which is expressed by the formula:

Where:

• $Q/t$ = Rate of heat transfer (Power in Watts)
• $k$ = Thermal conductivity ($0.6071 \, \text{Wm}^{-1}\text{K}^{-1}$)
• $A$ = Surface area ($620 \, \text{m}^2$)
• $T_2 - T_1$ = Temperature difference ($25^\circ\text{C} - 15^\circ\text{C} = 10 \, \text{K}$)
• $d$ = Depth/thickness ($1.5 \, \text{m}$)

Solution

Step 1: Identify the given values.

• $k = 0.6071 \, \text{Wm}^{-1}\text{K}^{-1}$
• $A = 620 \, \text{m}^2$
• $\Delta T = 25 - 15 = 10 \, \text{K}$
• $d = 1.5 \, \text{m}$

Step 2: Substitute the values into the formula.

Step 3: Perform the calculation.

• Numerator: $0.6071 \times 620 \times 10 = 3764.02$
• Divide by depth: $3764.02 / 1.5 = 2509.346 \, \text{W}$

Step 4: Convert to kilowatts (kW). Since $1000 \, \text{W} = 1 \, \text{kW}$:

Rounding to the nearest significant value provided in the options, we get 2.5 kW.

Conclusion

Comparing our result to the options provided, the calculated value matches option A.

Exam Tip

When solving heat transfer problems, always ensure your temperature difference is in Kelvin. However, because the formula uses the difference in temperature ($\Delta T$), a difference of $10^\circ\text{C}$ is exactly the same as a difference of $10 \, \text{K}$, so you don't need to convert the individual temperatures to Kelvin first!