How Much Natural Gas Is Needed to Heat a House to 28°C?

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To determine the amount of natural gas needed to heat a house to 28°C from an outside temperature of 0°C, the thermal conductivity of the house's walls and roof, along with their dimensions, must be accurately applied. The average thermal conductivity is 4.8 x 10^-4 kW/m°C, and the thickness of the walls is 20 cm. The calculation involves finding the power loss through the walls using the formula P = k * A * (ΔT / L), where A is the surface area and ΔT is the temperature difference. The total energy required over a day can then be calculated by multiplying the power by the number of seconds in a day. The discussion highlights confusion around the initial calculations and the proper application of thermal conductivity in determining heat loss.
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The average thermal conductivity of the walls (including windows) and roof of a house in Figure P11.32 is 4.8 10-4 kW/m°C, and their average thickness is 20.0 cm. The house is heated with natural gas, with a heat of combustion (energy given off per cubic meter of gas burned) of 9300 kcal/m3. How many cubic meters of gas must be burned each day to maintain an inside temperature of 28.0°C if the outside temperature is 0.0°C? Disregard radiation and loss by heat through the ground.

The house's dimensions are 10.0m x 8.0m x 5.0m


I tried this problem, and I keep getting the answer completely wrong... if anyone could show me how to do this one, that would be awesome.

Thanks.
 
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I did P=(4.8E-4)(.2)(28)

P= 2.688W


Q=Pnet(t)
Q=(2.688)(86400)
Q= 232243.2 J of heat


but i don't think that I am even starting it right.
 
Why did you multiply the thermal conductivity by length when computing the emitted power? How do you use the thermal conductivity to calculate the heat flux through a surface of area A and thickness L with a temperature difference \Delta T?
 
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