Question

Heat is lost at a rate of 275 W per m² area of a 15-cm-thick wall with a thermal conductivity of k=1.1W/m*K. The temperature drop across the wall is

A. 37.5 C
B. 4.0 C
C.27.5 C
D. 16.0 C
E. 8.0 C

Answer 1

The temperature drop across a 15-cm-thick wall with a thermal conductivity of 1.1W/m*K and a heat loss rate of 275 W per m² is 37.5 C.

Step-by-step explanation:

To determine the temperature drop across the wall, we can use the heat conduction equation which is Q = kAΔT/d, where Q is the heat transfer per unit time (Watts), k is the thermal conductivity (W/m*K), A is the area through which heat is being transferred (m²), ΔT is the temperature difference (Celsius or Kelvin, since they have the same increment), and d is the thickness of the wall (m).

In this case, we are given the following values: Q = 275 W/m², k = 1.1 W/m*K, and d = 0.15 m. We need to find ΔT. Rearranging the equation for ΔT, we get ΔT = Qd/(kA). Plugging in the values, ΔT = (275 W/m² * 0.15 m) / (1.1 W/m*K). This calculation results in a temperature drop of 37.5 C, so the correct answer is A: 37.5 C.