- #1
jdawg
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Homework Statement
The concrete slab of a basement is 11m long, 8m wide, and 0.20m thick. During the winter, temperatures are nominally% °C and 10°C at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of 1.4W/m*K, what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operating at an efficiency of 90% and natural gas is priced at Cg=$0.02/MJ, what is the daily cost of the heat loss?
Homework Equations
Simplification of Fourier's Law: q'' = -k(T1-T2)/L
q=q''*A
The Attempt at a Solution
So I had no problems calculating the heat loss (q=4312 W), but I'm a bit confused on how to incorporate the efficiency. I'm used to just multiplying the efficiency by the amount of energy that I calculated in order to find the actual energy produced or lost, but I know in this case I'm trying to calculate the cost.
This is the formula that the solution I have uses:
Cost = (q*t*Cg) / efficiency (Where q=heat loss, t=time,Cg=cost per Mg)
How do they know to divide by the efficiency?? I guess in the problem they tell you that the furnace giving off energy is 90% efficient and we are looking at the daily cost of heat loss and not heat produced...
I'm having a bit of a mental block with this problem, I would really appreciate an explanation :)