Question

A 2.5 m-high, 4 m-wide, and 20 cm-thick wall of a house has thermal resistance of 0.0125°C/W. The thermal conductivity of the wall is:

A. 0.72 W/m*K
B. 1.1 W/m*K
C. 1.6 W/m*K
D. 16 W/m*K
E. 32 W/m*K

Answer 1

The thermal conductivity of the wall is 0.00625 °C/(W*m²).

Step-by-step explanation:

The thermal conductivity of the wall can be determined using the formula:

Thermal Conductivity = Thermal Resistance / (Wall Thickness * Surface Area)

Given the values:

• Thermal Resistance = 0.0125 °C/W
• Wall Thickness = 20 cm = 0.2 m
• Surface Area = Height * Width = 2.5 m * 4 m = 10 m²

Plugging these values into the formula, we get:

Thermal Conductivity = 0.0125 °C/W / (0.2 m * 10 m²) = 0.0125 °C/W / 2 m² = 0.00625 °C/(W*m²)

Therefore, the thermal conductivity of the wall is 0.00625 °C/(W*m²).