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].

temperature gradient  = 35.78°C/m

Explanation:

given data

cross-sectional area A1 = 1.7 cm²

temperature gradient 1= 80 °C/m

cross-sectional area A2 = 3.8 cm²

to find out

temperature gradient   2

solution

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

Q = KA

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

A is cross section area and is temperature gradient

so for both Q will be equal

KA1 1 = KA2 2

put here value

1.7 × 80 = 3.8 ×   2

2 = 35.78

so temperature gradient  = 35.78°C/m

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