University of Wisconsin Green Bay

A 40 ft by 40 ft square house (with no basement) has outer walls that are 12 feet high. The roof rises to a height of 17 feet above the walls at the center, and the total area of windows and doors is 190 square feet. The house is insulated--the R-value of the insulation in the floor is 25 (ft2 0F hr/Btu), of the insulation in the roof is 55, and of the insulation in the walls is 15. Take the R value of the windows and doors to be about 5. The house is heated to 680F on a night when the outside air temperature is 200F and the windchill is 50F. If the house is well sealed against drafts and the furnace fails, about how long will it take for the inside temperature of the house to decrease to 650F? Take the ground temperature to be 500F, and assume there are no other sources of heat in the house.

  • At root, this is a heat transfer problem: you want to know how much time it takes place to lose enough heat for the house to cool 50.

    You know that insulation in a house reduces the heat transfer by conduction, and so even if you were not explicitly given the R-values you should recognize that you need to calculate heat transfer by conduction. The problem states that the house is well sealed, which implies that you do not need to take convection into account. The third method of heat transfer is radiation. In this case, you do not have any information about the emissivity of the house and therefore cannot calculate the rate of heal loss by radiation. It is certainly true that convection and radiation do play a role. The house will cool faster than the rate that you calculate from conduction alone.

    Note that you are not told in the problem how much thermal energy you need to transfer. This, then, is a two-part problem. Before you can find how much time it takes to transfer the energy, you need to find how much energy must leave the house as it cools 50F.

  • reference frames

    A picture of the house is not particularly useful. There is no loss to understanding if you skip this step. On the other hand, it might help you to visualize the information about the shape of the roof.

  • There are two versions of the relation that quantizes heat transfer by conduction. Because R-values were specified, I chose that version rather than Q/Δt = (A ΔT)/L. In this problem, you are not told how much energy is transferred. However, you are told that the temperature of the house falls and so you also need the equation that relates energy to temperature change.

  • The first step in this problem is to determine how much energy needs to leave the house in order for the temperature to drop 3 0F.


    Step 2

    Once you know how much energy needs to leave the house in order for it to cool to 65 0F, you can calculate how long it takes the energy to leave.

  • The house cools because thermal energy is transferred across all surfaces of the house. The rate at which thermal energy is transferred depends on the amount of insulation as well as the temperature difference across each surface. For that reason, the total heat transfer had to be calculated as the sum of the parts--transfer through the walls, through the windows, etc.

    Although physical constants could be looked up, a number of estimations were also made. For example, we assumed the entire house was filled with air and made assumptions about the shape of the roof. For that reason, it does not make sense to claim multiple significant figures.