Hello everybody! I am stuck in this problem (College Physics 7th Edition #32), a

Question

Hello everybody! I am stuck in this problem (College Physics 7th Edition #32), and... here's the outline of the problem

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 21.0 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 26.5°C if the outside temperature is 0.0°C? Disregard radiation and loss by heat through the ground.

(Note that the height is actually 5m, not 3m, and the angle is 37 degrees)

Before I go about asking for the solution, I figured out the surface area of the house, which was:

(8*5*2)+(10*5*2) = 180 m^2

Now... after that, I just can't seem to progress at ALL. I tried to find the surface area of the roof, but don't know what to do with the angle, and even if I DID find the area, I don't know how to proceed from there on. With thermal cunductivity, heat of combustion... I actually figured out the change of temperature (26.5, not hard to pull off at all).

Help is much appreciated

Explanation / Answer

to find the total area of the house. you have to find area of the roof. the angle 37 give you the height of the roof. h=8*sin37/2=2.4(m) so that area of the the roof. (2.4*8/2)*2+10*4*2/cos37=120(m2). so total area = 300. we have that ?P=C*S*?T/L=18171(W). where C is thermal conductivity. S is total area L is 21cm. ------------ from this we have in 1 day. E=18171*86400=1.57e9(J). so that V=1.57e9J/9300kcal=1.57e9/9300*4186=40.3(m3)

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