A house has well-insulated walls 17.5 cm thick (assume conductivity of air) and

Question

A house has well-insulated walls 17.5 cm thick (assume conductivity of air) and area 406 m2, a roof of wood 6.5 cm thick and area 274 m2, and uncovered windows 0.65 cm thick and total area 33 m2. Take the specific heat of air to be 0.24 kcal/kgC. (a) Assuming that the heat loss is only by conduction, calculate the rate at which heat must be supplied to this house to maintain its temperature at 23C if the outside temperature is -13C. kW (b) If the house is initially at 5C, estimate how much heat must be supplied to raise the temperature to 23C within 30 min. Assume that only the air needs to be heated and that its volume is 750 m3. J (c) If natural gas costs $0.080 per kilogram and its heat of combustion is 5.4 107 J/kg, how much is the monthly cost to maintain the house as in part (a) for 24 h each day assuming 93 percent of the heat produced is used to heat the house? dollars/month

Explanation / Answer

The initial temperature difference is 23 C°. The thermal conductance through the walls is (0.024 J/s·m·C°)(23 C°)(406 m²) / (0.065 m) = 3447.8 W (to 3 significant figures). The thermal conductance through the roof is (0.1 J/s·m·C°)(23 C°)(272 m²) / (0.065 m) = 9621 W (to 3 significant figures). The thermal conductance through the windows is (0.84 J/s·m·C°)(23 C°)(33 m²) / (0.065 m) = 8530 W (to 3 significant figures). The total thermal conductance (heat loss) is 21 598 W (3 sig figs). The specific heat capacity of air in the given temperature range is 1005 J/kg/K; the density is 1.25 kg/m³, so the mass is (750 m³)(1.25 kg/m³) = 938 kg. To raise the temperature by 13 C° requires (1005 J/kg/K)(938 kg)(13 K) = 12.25 MJ. While the house is being heated, though, 18 200 J/s are being lost through the structure; in 30 min, this is 32.76 MJ. So, a total of 12.25+32.76 = 45.0 MJ will need to be put into the house.

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