Question

Suppose that the rod in the figure is made of copper (for copper, k = 385 W/m∙K), is 45.0 cm long, and has a cross-sectional area of 1.25 cm2. Let TH = 100 °C and TC = 0.0 °C. (a) What is the final steady-state temperature gradient along the rod?

I forget theory and formulas (. May be someone give me a hint ?

Answer 1

This question totally stumped me. I figured there would be a minus sign in front of the equation. I went to dig up my old notes, and found that the equation is:
\[\frac{dQ}{dt}=-kA\frac{dT}{dx}\]
You will not need this for part a though.

Assuming Tc and Th doesn't change, then the steady state equation is:
\[\frac{dT}{dx}=\frac{(T_{2}-T_{1})}{(x_{2}-x_{1})}=100/0.45=222Km^{-1}\]

Answer 2

You must use the Heat equation that goes:

\(\LARGE \frac {\partial ^2 T}{\partial x^2}-\frac {\rho c}{k}\frac {\partial T}{\partial t}\large =0\)

In steady state, it simplifies to

\(\LARGE \frac {d ^2 T}{d x^2} =0\)

which is extremely easy to integrate and will lead to the equation given by Festinger.