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InterviewSolution
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A wall of dimensions 2.00 m by `3.50 m` has a single-pane window of dimensions `0.75 m` by `1.20 m` . If the inside temperature is `20^(@)C` and the outside temperature is `-10^(@)C` , effective thermal resistance of the opaque wall and window are `2.10 m^(2) K W^(-1)` and `0.21 m^(2) K W^(-1)` respectively. The heat flow through the entire wall will be .A. `215 W`B. `205 W`C. `175 W`D. `110 W` |
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Answer» Correct Answer - A The wall and the window are in parallel arrangement , so net heat flow is the sum of heat flow through the wall and the window. The temperature difference, `T_(H) - T_(C) = 30 K` Area of window, `A_(i) = (0.75) (1.20) = 0.90 m^(2)` Heat flow through window pane, `((dQ)/dt)_(1)= (T_(H) - T_(C))/(R_(1)) = (0.90)(30)/(0.21) = 128.6 W` The area of the wall, `A_(2) = (2.00)(3.50) - (0.75)(1.20) = 6.10 m^(2)` Heat flow through the wall, `((dQ)/dt)_(2) = (T_(H) - T_(C))/(R_(2)) = ((6.10)(30))/((2.10)) = 87 W` Net heat flo, `(dQ)/(dt) = 128.6 + 87 = 215.6 W` |
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