Question

Given a thermal conductivity constant of a material of 0 4, a surface area of 100 square feet, a temperature on the far side of a material of 165°F and a temperature of 150°F on the near side, and a material thickness of 1.5 inches, what is the conductive heat transfer rate? 2400 F per ft. 4800 F per ft. 3600 F per ft. 6400 F per ft.

Answer 1

Answer:4800

Step-by-step explanation:

Given data

k=0.4

surface area(A)=100
`ft&^(2)`

Temprature on near side of material
`(T_1)`=150°F

temprature on far side of material
`(T_2)`=165°F

thickness(t)=1.5in=0.125
`ft^2`

Conductive heat transfer rate(Q) =
`(k* A*\left ( T_2-T_1\right ))/(y)`

Q=
`(0.4* 100\left ( 165-150\right ))/(1.5)`

Q=4800°F per feet