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

`(dQ)/(dt) = 966 W`

Step-by-step explanation:

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

`(dQ)/(dt) = (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

`(dQ)/(dt) = ((0.69W/m-K)(28 m^2)(15))/(0.30)`

`(dQ)/(dt) = 966 W`