Question

Heat is lost at a rate of \(275 \mathrm{~W}\) per \(\mathrm{m}^{2}\) area of a 15 -cm-thick wall with a thermal conductivity of $k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The temperature drop across the wall is (a) \(37.5^{\circ} \mathrm{C}\) (b) \(27.5^{\circ} \mathrm{C}\) (c) \(16.0^{\circ} \mathrm{C}\) (d) \(8.0^{\circ} \mathrm{C}\) (e) \(4.0^{\circ} \mathrm{C}\)

Short answer

Answer: (a) 37.5°C

Step-by-step solution

Write down the Fourier's law of conductive heat transfer formula

The formula for Fourier's law of conductive heat transfer is: \(Q = -kA \frac{dT}{dx}\), where \(Q\) is the heat transfer rate, \(k\) is the thermal conductivity, \(A\) is the area of the wall, \(dT\) is the temperature difference across the wall (temperature drop), \(dx\) is the thickness of the wall. Since we already know the heat transfer rate per unit area (Q/A), we can rewrite the formula as: \(\frac{Q}{A} = -k \frac{dT}{dx}\)

Convert all units to SI units

To make all the calculations easier, convert the thickness of the wall from centimeters to meters: Thickness of the wall in meters: \(15 \mathrm{~cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.15 \mathrm{~m}\)

Plug in the given values into the formula

Now, we can plug the given values to the equation: \(\frac{275 \mathrm{~W}}{\mathrm{m^2}} = 1.1 \mathrm{~W/(m \cdot K)} \frac{dT}{0.15 \mathrm{~m}}\)

Solve for the temperature drop (ΔT)

Now, let's solve the equation for the temperature drop, \(dT\): \(dT = \frac{275 \mathrm{~W} \cdot 0.15 \mathrm{~m}}{1.1 \mathrm{~W/(m \cdot K)} \cdot \mathrm{m^2}}\) \(dT \approx 37.5 \mathrm{~K}\) Since we can interpret a temperature difference in Kelvins as the same temperature difference in Celsius, we can also write the temperature drop as \(37.5^{\circ} \mathrm{C}\). So, the correct answer is (a) \(37.5^{\circ} \mathrm{C}\).