Question

. The inner and outer surfaces of a 4-m × 7-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m-K are maintained at temperatures of 20 °C and 5 °C, respectively. Determine the rate of heat transfer through the wall, in W.

Answer 1

Answer:

$\frac{dQ}{dt} = 966 W$

Explanation:

As we know that the rate of heat transfer due to temperature difference is given by the formula

$\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}$

here we know that

$K = 0.69 W/m-K$

A = 4 m x 7 m

thickness = 30 cm

temperature difference is given as

$\Delta T = 20 - 5 = 15 ^oC$

now we have

$\frac{dQ}{dt} = \frac{(0.69W/m-K)(28 m^2)(15)}{0.30}$

$\frac{dQ}{dt} = 966 W$