contestada

. 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.

Respuesta :

Answer:

[tex]\frac{dQ}{dt} = 966 W[/tex]

Explanation:

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

[tex]\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}[/tex]

here we know that

[tex]K = 0.69 W/m-K[/tex]

A = 4 m x 7 m

thickness = 30 cm

temperature difference is given as

[tex]\Delta T = 20 - 5 = 15 ^oC[/tex]

now we have

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

[tex]\frac{dQ}{dt} = 966 W[/tex]

Otras preguntas