Since the rate of heat transfer by conduction is proportional to the $\Delta$T, what percent savings in your heating bill would you expect if you turned the thermostat down from 75 $\degree$ to 65 $\degree$ , and the outside temperature is 25 $\degree$F?
A) 10%
B) 20%
C) 30%
D) 40%
E) 50%

Question

Quizplus · Accepted Answer

The rate of heat transfer by conduction can be represented as:
Q = kA (T1 - T2)/d
where Q is the rate of heat transfer, k is the thermal conductivity, A is the surface area, T1 and T2 are the temperatures of the two sides of the material, and d is the thickness of the material.

Assuming that the surface area and thickness of the walls remain constant, the rate of heat transfer is directly proportional to the temperature difference (T1 - T2) between the inside and outside of the building.

Let's assume that the rate of heat transfer at 75°F is Q1, and the rate of heat transfer at 65°F is Q2. Then:
Q1/Q2 = (T1 - 75°F)/(T1 - 65°F)
Q2/Q1 = (T1 - 65°F)/(T1 - 75°F)

Assuming that the wall is made of the same material, the thermal conductivity is constant, so k cancels out. Therefore, the ratio of heat transfer rates depends only on the temperature difference.

Substituting the given values, we have:
Q1/Q2 = (T1 - 75°F)/(T1 - 65°F) = (25°F)/(15°F) = 5/3

This means that the rate of heat loss at 75°F is 5/3 times the rate of heat loss at 65°F. In other words, by turning down the thermostat from 75°F to 65°F, we can reduce the rate of heat loss by 3/5 or 60%, which corresponds to about a 20% savings in heating costs (since heat loss is proportional to heating costs). Therefore, the answer is B) 20%.

Since the Rate of Heat Transfer by Conduction Is Proportional Δ\DeltaΔ

Since the Rate of Heat Transfer by Conduction Is Proportional $\Delta$