Question

The roof of an electrically heated house is \(7-\mathrm{m}\) long, 10-m wide, and 0.25-m thick. It is made of a flat layer of concrete whose thermal conductivity is \(0.92 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C} .\) During a certain winter night, the temperatures of the inner and outer surfaces of the roof are measured to be \(15^{\circ} \mathrm{C}\) and \(4^{\circ} \mathrm{C}\) respectively. The average rate of heat loss through the roof that night was \((a) 41 \mathrm{W}\) \((b) 177 \mathrm{W}\) \((c) 4894 \mathrm{W}\) \((d) 5567 \mathrm{W}\) \((e) 2834 \mathrm{W}\)

Short answer

Answer: (e) 2834 W

Step-by-step solution

Understand the formula for heat transfer rate

The formula for heat transfer rate (Q) through a homogeneous material with constant thermal conductivity (k) is given by the equation: Q = k × A × (T1 - T2) / d where A is the area of the material, T1 and T2 are the temperatures on the two sides of the material, and d is the thickness of the material. In this problem, we have the thermal conductivity (k), dimensions of the roof, and the temperature difference (T1 - T2). So, we can use this formula to calculate the heat transfer rate.

Calculate the area of the roof

The area of the roof can be calculated using the formula for the area of a rectangle: A = Length × Width A = \((7\,\text{m}) (10\,\text{m})\) A = \(70\,\text{m}^2\)

Calculate the temperature difference

The temperature difference between the inner and outer surfaces of the roof is given by the equation: ΔT = T1 - T2 ΔT = \((15^{\circ}\,\text{C}) - (4^{\circ}\,\text{C})\) ΔT = \(11^{\circ}\,\text{C}\)

Calculate the heat transfer rate

Now we have all the information needed to calculate the heat transfer rate using the formula mentioned in step 1. Plug in the values into the formula: Q = \((0.92\,\frac{\text{W}}{\text{m}\cdot^{\circ}\mathrm{C}}) \times (70\,\text{m}^2) \times \frac{(11^{\circ}\,\text{C})}{(0.25\,\text{m})}\) Q = \(2831.6\,\text{W}\)

Compare the result with the given options

Compare the calculated heat transfer rate with the given options to find the closest match: (a) 41 W (b) 177 W (c) 4894 W (d) 5567 W (e) 2834 W The closest match to our calculated heat transfer rate of 2831.6 W is the value \((e)\,2834\,\text{W}\), which is approximately the same. So, the average rate of heat loss through the roof that night was \((e)\,2834\,\text{W}\).