 In this problem, we are trying to calculate, again, heat loss through a heating season. We are given the area. Area of the wall is 160 foot square. And we are given the heating degree days. Heating degree days for the season. This is, again, state college. Therefore, we can take it as 6,000 degrees days. Now we need R. The wall appears to consist of four layers. One is the brick wall. And this brick wall is 3 inches thick. And each inch gives us an R value of 0.2. So 3 inches would offer a 0.6 R value. And similarly, 12 inches of cinder block. Each inch has an R value of 1.89. Therefore, 12 inches would provide us 22.68. We have 2 inches of fiberglass. And fiberglass provides per inch an R value of 3.7. Therefore, 2 inches would be 7.40. And now we have finally half inch drywall. And the whole half inch of drywall will provide us an R value of 0.45. So together, when you add all these layers up, the R value of the wall happens to be 31.13. Now heat loss through a season is given by area times HDD times 24 over R. So in this case, it happens to be 160 foot square times 6,000 degree days times 24 hours over a day divided by we have an R value of 31.13 foot square degrees Fahrenheit hour over BTU. Now foot square, foot square can be canceled. Degrees Fahrenheit, degrees Fahrenheit, days and days, hours and hours are canceled here. And we get 740,122 BTUs.