Question

The north wall of an electrically heated home is \(20 \mathrm{ft}\) long, $10 \mathrm{ft}\( high, and \)1 \mathrm{ft}$ thick and is made of brick whose thermal conductivity is $k=0.42 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}$. On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about \(62^{\circ} \mathrm{F}\) and \(25^{\circ} \mathrm{F}\), respectively, for a period of \(8 \mathrm{~h}\). Determine \((a)\) the rate of heat loss through the wall that night and \((b)\) the cost of that heat loss to the homeowner if the cost of electricity is \(\$ 0.07 / \mathrm{kWh}\).

Short answer

Answer: The rate of heat loss through the brick wall that night is 3108 Btu/h, and the cost of that heat loss to the homeowner is $0.51.

Step-by-step solution

Gather given information

We are given the following information: - Wall length (L) = 20 ft - Wall height (H) = 10 ft - Wall thickness (t) = 1 ft - Thermal conductivity (k) = 0.42 Btu/h·ft·°F - Inner wall temperature (T1) = 62°F - Outer wall temperature (T2) = 25°F - Electricity cost = $0.07/kWh - Time period (time) = 8 hours

Calculate the temperature difference

The first step in determining the rate of heat loss is to calculate the temperature difference between the inner and outer surfaces of the wall. This can be calculated as: \(\Delta T = T1 - T2 = 62 - 25 = 37^{\circ} \mathrm{F}\)

Calculate the area of the wall

Next, we need to find the area of the wall. Since the wall is a rectangle, we can calculate the area as: \(A = L \times H = 20 \times 10 = 200 \mathrm{ft}^2\)

Apply Fourier's law of heat conduction

Fourier's law of heat conduction states that the heat loss rate (Q) can be found using the formula: \(Q = \frac{kA \Delta T}{t}\) Inserting our known values, we get: \(Q = \frac{0.42 \cdot 200 \cdot 37}{1} = 3108 \ \mathrm{Btu/h}\) So, the rate of heat loss through the wall that night is 3108 Btu/h.

Convert heat loss rate to kWh

To find the cost, we need to convert the heat loss rate in Btu/h to energy in kilowatt-hours (kWh). Using the conversion factor (1 kWh = 3412 Btu), we can calculate the total energy lost in kWh over the 8-hour period: \(E = \frac{Q \cdot time}{3412} = \frac{3108 \cdot 8}{3412} = 7.29 \ \mathrm{kWh}\)

Calculate the cost of heat loss

Finally, using the given cost of electricity (\(0.07 / \mathrm{kWh}\)), we can find the cost of heat loss to the homeowner: \(Cost = E \cdot rate = 7.29 \cdot 0.07 = \$0.51\) Thus, \((a)\) the rate of heat loss through the wall that night is 3108 Btu/h, and \((b)\) the cost of that heat loss to the homeowner is $0.51.