(a) Calculate the heat flux through a sheet of brass \(7.5 \mathrm{mm}(0.30 \text { in. })\) thick if the temperatures at the two faces are 150 and \(50^{\circ} \mathrm{C}\) \(\left(302 \text { and } 122^{\circ} \mathrm{F}\right) ;\) assume steady-state heat flow. (b) What is the heat loss per hour if the area of the sheet is \(0.5 \mathrm{m}^{2}\left(5.4 \mathrm{ft}^{2}\right) ?(\mathrm{c}) \mathrm{What}\) will be the heat loss per hour if soda-lime glass instead of brass is used? (d) Calculate the heat loss per hour if brass is used and the thickness is increased to \(15 \mathrm{mm}(0.59 \text { in. })\).

The formula for heat conduction is given by: $$q = \frac{k A \Delta T}{d}$$ Where: - \(q\) is the heat flux (W) - \(k\) is the thermal conductivity of the material (W/mK) - \(A\) is the area through which the heat is conducted (m^2) - \(\Delta T\) is the temperature difference between the two sides of the material (K) - \(d\) is the thickness of the material (m)

We are given the following values for the brass sheet: - \(d = 7.5\) mm = 0.0075 m (Convert mm to m) - \(T_1 = 150^\circ\)C - \(T_2 = 50^\circ\)C - \(k_\text{brass} = 109\) W/mK (Thermal conductivity of brass) Now, let's apply the heat conduction formula: $$q_\text{brass} = \frac{k_\text{brass} \cdot \Delta T_\text{brass}}{d_\text{brass}}$$ Where: - \(\Delta T_\text{brass} = T_1 - T_2 = 150 - 50 = 100\) K - \(d_\text{brass} = 0.0075\) m Plug in the provided values into the formula: $$q_\text{brass} = \frac{109 \cdot 100}{0.0075} = 1,453,333.3 \thinspace \text{W} $$

We have already calculated the heat flux (\(q_\text{brass}\)) in Step 2. Now we need to find the heat loss per hour. We are given the area of the brass sheet, \(A = 0.5 m^2\). The formula for heat loss (\(Q\)) is as follows: $$Q = q \cdot A \cdot t$$ Where \(t\) is the time in hours. In this case, we want to calculate the heat loss per hour, so \(t = 1 \thinspace \text{hour}\). $$Q_\text{brass} = q_\text{brass} \cdot A \cdot 1 = 1,453,333.3 \cdot 0.5 \thinspace \text{W} = 726,666.6 \thinspace \text{W hour}$$

The heat loss per hour equation remains the same, but we need to use the thermal conductivity of soda-lime glass (\(k_\text{glass} = 1.4 \text{ W/mK}\)) and the same thickness and temperature difference as the brass sheet: $$q_\text{glass} = \frac{k_\text{glass} \cdot \Delta T_\text{brass}}{d_\text{brass}} = \frac{1.4 \cdot 100}{0.0075} = 18,666.7 \thinspace \text{W}$$ Now we can find the heat loss per hour for the glass sheet: $$Q_\text{glass} = q_\text{glass} \cdot A \cdot 1 = 18,666.7 \cdot 0.5 \thinspace \text{W} = 9,333.3 \thinspace \text{W hour}$$

In this case, we need to use the new thickness provided for the brass sheet (\(d_\text{thick} = 15\) mm = 0.015$ m) and the same temperature difference, area, and thermal conductivity as before: $$q_\text{thick} = \frac{k_\text{brass} \cdot \Delta T_\text{brass}}{d_\text{thick}} = \frac{109 \cdot 100}{0.015} = 726,666.7 \thinspace \text{W}$$ Now we can find the heat loss per hour for the thicker brass sheet: $$Q_\text{thick} = q_\text{thick} \cdot A \cdot 1 = 726,666.7 \cdot 0.5 \thinspace \text{W} = 363,333.3 \thinspace \text{W hour}$$ So, the final answers are: (a) Heat flux for brass: \(1,453,333.3 \thinspace \text{W}\) (b) Heat loss per hour for brass: \(726,666.6 \thinspace \text{W hour}\) (c) Heat loss per hour for soda-lime glass: \(9,333.3 \thinspace \text{W hour}\) (d) Heat loss per hour for thicker brass: \(363,333.3 \thinspace \text{W hour}\)