If only conductive heat loss was significant for a house, then by what percentage would you lower the heat loss if the temperature was reduced inside from 70 F to 60 F when the outside temperature is 20 F?

Question

anonymous · Answer

Do you know the thermal conductivity of the wall?

anonymous · Answer

no other infor is given

anonymous · Answer

I guess not. Let's try and solve this using the heat conductivity equation.$$ Q = -kA(\Delta T)/(\Delta x)$$

Where Q represents heat loss, k represents thermal conductivity, A represents area of the wall, and x is the thickness of the wall.

We'll assume everything except Q and delta T is for this case only, 1.

Case 1 : 70 F - 20 F = 50 F.
$$Q=−1\times 1(50)/(1)$$
That's -50. Negative because energy is being transferred outside.
Assuming the same trend..

Case 2 : 60F - 20F = 40F difference.
I'm going to skip the calc. for this one, since it's pretty obvious. You're ending up with -40 as the answer here.

Now, let's express the answer of Case 2 where the temp difference is -40, as a percentage of Case 1. 

$$\frac{ -40 }{-50} \times 100$$$$= 80%$$

Therefore, you're lowering the heat loss by 20%. 

The reason I included all the steps with the equation is just to make things clear. Otherwise, you could just have used the unitary method in one step to find the answer.

Hope this helped you! -swas

anonymous · Answer

This makes sense! I was thrown off by not having a specific area or thickness of a wall to work with

anonymous · Answer

Glad I could be of help. If you haven't already, close the question :)