QuestionDecember 14, 2025

33. II A 10mtimes 14m house is built on a 12-cm-thick concrete slab. What is the heat-loss rate through the slab if the ground tempera- ture is 5^circ C while the interior of the house is 22^circ C ?

33. II A 10mtimes 14m house is built on a 12-cm-thick concrete slab. What is the heat-loss rate through the slab if the ground tempera- ture is 5^circ C while the interior of the house is 22^circ C ?
33. II A 10mtimes 14m house is built on a 12-cm-thick concrete slab. What is the heat-loss rate through the slab if the ground tempera- ture is 5^circ C while the interior of the house is 22^circ C ?

Solution
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Answer

27,\!800\,\text{W} (rounded to three significant digits) Explanation 1. Calculate the area of the slab A = 10\,\text{m} \times 14\,\text{m} = 140\,\text{m}^2 2. Identify thickness and temperature difference Thickness L = 12\,\text{cm} = 0.12\,\text{m}; \Delta T = 22^\circ C - 5^\circ C = 17\,\text{K} 3. Use thermal conductivity for concrete k_{\text{concrete}} \approx 1.4\,\text{W}/(\text{m}\cdot\text{K}) 4. Apply heat-loss rate formula **Q = \frac{kA\Delta T}{L}** 5. Substitute values and calculate Q = \frac{1.4 \times 140 \times 17}{0.12} = \frac{3332}{0.12} \approx 27,\!767\,\text{W}

Explanation

1. Calculate the area of the slab <br /> $A = 10\,\text{m} \times 14\,\text{m} = 140\,\text{m}^2$<br />2. Identify thickness and temperature difference <br /> Thickness $L = 12\,\text{cm} = 0.12\,\text{m}$; $\Delta T = 22^\circ C - 5^\circ C = 17\,\text{K}$<br />3. Use thermal conductivity for concrete <br /> $k_{\text{concrete}} \approx 1.4\,\text{W}/(\text{m}\cdot\text{K})$<br />4. Apply heat-loss rate formula <br /> **$Q = \frac{kA\Delta T}{L}$**<br />5. Substitute values and calculate <br /> $Q = \frac{1.4 \times 140 \times 17}{0.12} = \frac{3332}{0.12} \approx 27,\!767\,\text{W}$
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