Heat equation and energy transport.

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SUMMARY

The discussion focuses on solving the heat equation for a rod with uniform density \(\rho\) and length \(l\), where the internal energy density is defined as \(E = c(T - T_0) + E_0\). The time-independent solution to the heat equation \(\frac{\partial T}{\partial t} = \kappa \nabla^2 T\) is derived as \(T(x) = \frac{T_l - T_0}{l}x + T_0\). The energy transport per unit time at the end of the rod is calculated using Fourier's Law, resulting in the expression \(q'' = \rho c \kappa \left(\frac{T_0 - T_l}{l}\right)\). The total thermal energy in the rod is determined through integration, yielding \(E = \rho S l \left[\frac{c (T_l - T_0)}{2} + E_0\right]\).

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Homework Statement


I have a rod of density \rho and length l. It's located at 0\leq x\leq l. The density of internal energy per mass is E = c(T-T_0) + E_0 where T is the tempertature in Kelvin,E_0 is a constant and c is the specific heat capacity. We assume that the temperature is not varying across the rodd. The temperature at the two ends of the rod is T_0, T_l

a) this question was to find the time independet solution to the heatequation
\frac{\partial T}{\partial t} = \kappa \nabla^2 T and I found this one by using the conditions to be

T(x) = \frac{T_l - T_0}{l}x + T_0

b) Find the transport of energy, per unit time, out of a cross-section of the rod at x = l.Also find the total thermal energy in the rod.

Homework Equations




The Attempt at a Solution


Im thinking that the solution to this probably is a flux integral, but I don't know how to proceed and what to integrate.
 
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Transport of energy, per unit time, out of a cross-section of the rod. That sounds like heat flux. We use Fourier's Law in order to obtain the expression for heat flux.
q'' = -k \frac{dT}{dx} = - \rho c \kappa \frac{dT}{dx}
q'' = \rho c \kappa \left(\frac{T_0 - T_l}{l} \right)
Now, in order to find the total thermal energy of the rod we must integrate the energy density with respect to mass
E = \int_0^m \hat{E} \ dm
Let S be the cross-section area of the rod, then
dm = \rho S dx
E = \rho S \int_0^l \left[c (T_l - T_0) \frac{x}{l} + E_0 \right] dx
E = \rho S l \left[ \frac{c (T_l - T_0)}{2} + E_0 \right]
 

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