Question

One easy way to reduce heating (and cooling) costs is to add extra insulation in the attic of a house. Suppose the house already had 15 cm of fiberglass insulation in the attic and in all the exterior surfaces. If you added an extra 8.0 cm of fiberglass to the attic, then by what percentage would the heating cost of the house drop? Take the single story house to be of dimensions 10 m by 15 m by 3.0 m. Ignore air infiltration and heat loss through windows and doors.

The cost of heating would drop by 12%.

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This is College Physics Answers with Shaun Dychko. In this question, we consider a house with dimensions of 10 meters length, 15 meters width and 3 meters height and there is insulation on each of these surfaces, these walls, the floor and the attic ceiling and the insulation thickness is 15 centimetres which is 15 times 10 to the minus 2 meters and we are meant to add a delta

*d*increase the thickness of the insulation in the attic only by 8 centimetres. Now, the thermal conductivity of fibre glass insulation is, is called glass wool in our data table; it's kind of a funny way of calling it but it does make sense. It's made out of glass and it is like wool. And it's 0.042 joules per second per meter per Celsius degree and the question's asking us what is the percent change in the heating cost of this house when you add 8 centimetres of insulation to the attic. Now, heating cost is proportional to the rate of heat transfer due to conduction and so that means there is some number, we'll call it*z*; doesn’t matter what it is, multiplied by the rate of thermal conduction that will turn this rate into a cost in dollars and cents and so percent change in cost will be the cost 2 after the insulation is added to the attic minus the initial cost divided by the initial cost times 100 percent and then we can substitute*z**Q*over*t*in place of each of those costs, having a subscript 2 for the cost after adding insulation and cost 1 or*Q 1*for the cost before adding insulation and you see that this*z*is a common factor and so it doesn’t matter what it is because it's asking for the percent change, we can ignore that constant of proportionality and so when we divide the denominator into the numerator, the*z*'s cancel and we are left with 1 here because this is the same and then on the left, we have*Q 2*over*t*divided by*Q 1*over*t*. So, we are gonna figure out this ratio minus 1 times 100 percent to get the percent change in the cost. Now, our formula for the initial rate of heat conduction is going to be the thermal conductivity of the glass wool multiplied by the total area times the difference in temperature divided by the thickness and so the total area would be 2 times this surface here*A 1*coz there's this one that we can see and then there's one at the back as well, that wall, and then plus 2 times area 2 this wall plus the wall that is on this side as well on the ends and then plus 2 times*A 3*so the attic and plus the floor and remember this is when the attic and floor have the same thickness and so that’s why you can just put them together inside this bracket here whereas down here we have to separate it out later when we talk about the*Q final*rate of heat conduction after adding the insulation. So, by the way up here, I have*Q 2*over*t*over*Q 1*over*t*I switched to using letters initial and final because the subscripts 1, 2 and 3 are getting confused with the labels I have put on these sides of the house. So let's change this to*Q final*and then initial. Okay. I think you get the idea and then let's go back here. Alright, then we need to know what each of the areas are. So we have*Area 1*is 10 meters by 3 meters so that’s 30 square meters.*Area 2*is 15 meters by 3 meters, which is 45 square meters. Then,*Area 3*is 10 meters by 15 meters for an area of 150 square meters and then the initial rate of heat conduction then is going to be over factoring out the 2 so that’s where this 2 comes from and then multiplying by the thermal conductivity of glass wool then multiplying by each of these areas added together times delta*T*over the thickness of 15 centimetres which is 15 times 10 to the minus 2 meters. This gives 126 watts per Celsius degree of heat of temperature difference multiplied by the temperature difference and we are not getting an answer as a single number, we are getting answer in terms of delta*T*. But when we divide, we are gonna see that this delta*T*cancels because we are gonna divide this initial into the final. So now let's also calculate the rate of heat transferred due to conduction after adding the insulation to the attic and so we have this term which is all of the sides which have the original initial thickness of insulation and notice the*A 3*is multiplied just by 1 not 2 because it's only the floor that has the initial thickness here and whereas the attic now has its separate term, it's gonna be*k*times*A 3*area of the ceiling or attic, whatever you wanna call it, times delta*t*over a new final thickness and that final thickness is going to be the initial thickness plus what we are adding so it’s 15 centimetres plus the 8 centimetres added and I factored out the*k*delta*T*from each of these terms and wrote those here and so we have*k*delta*T*times 2*A 1*plus 2*A 2*plus*A 3*all over initial thickness plus*A 3*over the final thickness which is initial plus the additional added. Okay, so let me plug in numbers. So, that’s 0.042 joules per second per meter per Celsius degree times the temperature difference, whatever that is, times 2 times 30 square meters plus 2 times 45 square meters plus the floor area 150 square meters divided by 15 centimetres plus the ceiling area divided by the 15 centimetres plus 8 centimetres, all written in meters, of course, and this gives 111.391 watts per Celsius degree times the temperature difference. Okay so the change in cost then is the, umm… is the final rate of heat transfer divided by initial rate of heat transfer minus 1 times 100 percent. So that’s final which is 111.391 watts per Celsius degree times change in temperature divided by the initial rate of heat transfer 126 watts per Celsius degree times the temperature difference minus 1 times a 100 percent and these cancel and all this works out to minus 12 percent. So, there is a 12 percent reduction in cost by adding 8 centimetres of insulation to the attic.