Question

In one section of a material, the cross-sectional area is 1.7 cm^2 and the temperature gradient is 80 °C/m. If the same rate of heat transfer occurs in the same material but at a location where the cross-sectional area is 3.8 cm2, calculte the temperature gradient at that location [°C/m].

Answer 1

temperature gradient = 35.78°C/m

Step-by-step explanation:

given data

cross-sectional area A1 = 1.7 cm²

temperature gradient
`(dT)/(dx)` 1= 80 °C/m

cross-sectional area A2 = 3.8 cm²

to find out

temperature gradient
`(dT)/(dx)` 2

solution

we will apply here fourier law that is for absolute value of temperature gradient

Q = KA
`(dT)/(dx)`

here Q is rate of heat transfer and K is conductivity of material

A is cross section area and
`(dT)/(dx)` is temperature gradient

so for both Q will be equal

KA1
`(dT)/(dx)` 1 = KA2
`(dT)/(dx)` 2

put here value

1.7 × 80 = 3.8 ×
`(dT)/(dx)` 2

`(dT)/(dx)` 2 = 35.78

so temperature gradient = 35.78°C/m