Efficiency Associated with Cost

[jdawg]

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 :)

 

[NTL2009]

Well, since they are looking for cost, consider how much would it cost to heat that space with a 50% efficient furnace, versus a (theoretical) 100% efficient furnace.

Does that explain it?

OK, mathematically, looking at the formula for cost, "efficiency" is in the denominator. So that mean cost goes down as efficiency goes up, right? Doesn't that make sense?

 

[jdawg]

Thank you so much! It makes total sense, great explanation!