Answer :
Answer:
[tex] k = - \frac{40 W/m^2}{\frac{20-40 C}{0.05 m}}= 0.1 \frac{W}{m C}[/tex]
Step-by-step explanation:
Previous concepts
The Fourier's law, states that the "rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area, at right angles to that gradient, through which the heat flows".
And we have the following formula given:
[tex] q_x = -k A \frac{dT}{dx}[/tex]
For this case we have the following data given:
[tex] L= 50 mm * \frac{1m}{1000 mm}=0.05m[/tex]
[tex] T_i = 40 C[/tex] represent the inner surface temperature
[tex] T_o = 20 C[/tex] represent the outer surface temperature
[tex] \frac{q_x}{A} = 40 W/m^2 [/tex] represent the heat flux
So then we are interested on the value of k.
Solution to the problem
if we solve for the value of k from the formula:
[tex] q_x = -k A \frac{dT}{dx}[/tex]
We got:
[tex] k = - \frac{q_x}{A \frac{dT}{dx}}[/tex]
We can rewrite this expression like this:
[tex] k = - \frac{q_x}{A \frac{T_o -T_i}{L}}[/tex]
And if we replace the values given we got:
[tex] k = - \frac{40 W/m^2}{\frac{20-40 C}{0.05 m}}= 0.1 \frac{W}{m C}[/tex]