1. **State the problem:** Calculate the heat flow per hour through a brick and mortar wall 9 inches thick, 10 feet high, and 6 feet wide. The thermal conductivity coefficient is 500 $\frac{ft\cdot lb}{hr\cdot ft \cdot ^\circ C}$, with temperatures 330°F and 130°F on each side. 2. **Formula used:** Heat flow rate $Q$ is given by Fourier's law for heat conduction: $$Q = \frac{k A \Delta T}{L}$$ where: - $k$ is the thermal conductivity, - $A$ is the cross-sectional area, - $\Delta T$ is the temperature difference, - $L$ is the thickness of the wall. 3. **Convert units:** - Thickness $L = 9$ inches = $\frac{9}{12} = 0.75$ feet. - Area $A = 10 \times 6 = 60$ square feet. - Temperature difference $\Delta T = 330 - 130 = 200$ °F. 4. **Convert temperature difference to Celsius:** Since $1 ^\circ C = \frac{9}{5} ^\circ F$, $$\Delta T_C = \frac{5}{9} \times 200 = \frac{1000}{9} \approx 111.11 ^\circ C$$ 5. **Calculate heat flow $Q$:** $$Q = \frac{500 \times 60 \times 111.11}{0.75}$$ 6. **Simplify step-by-step:** $$Q = \frac{500 \times 60 \times 111.11}{0.75} = 500 \times 60 \times \frac{111.11}{0.75}$$ 7. **Calculate $\frac{111.11}{0.75}$:** $$\frac{111.11}{0.75} = \frac{\cancel{111.11}}{\cancel{0.75}} = 148.148$$ 8. **Calculate total heat flow:** $$Q = 500 \times 60 \times 148.148 = 500 \times 8888.88 = 4444440$$ 9. **Convert $ft\cdot lb$ to Btu:** Given 1 Btu = 778 ft-lb, $$Q_{Btu/hr} = \frac{4444440}{778} \approx 5713.5$$ 10. **Final answer:** Approximately 5700 Btu/hr, which is closest to option c) 5400 Btu/hr. **Answer:** c) 5400 $\frac{Btu}{hr}$