Now we calculate the total heat loss for a rod with two heat sources. This is not very much different at all from our previous calculation. However, this time instead of just taking the heat flux from the two heat sources we will take the method of integrating the heat loss over the whole rod.

`> `
**restart;**

The general temperature distribution equation that we found previously, there is no need to recalculate it out.

`> `
**Theta:=x->Theta0*(exp(m*2*L-m*x)+exp(m*x))/(1+exp(m*2*L));**

Integrate for Heat loss over the entire Rod.

`> `
**eqn:=2*q[0]=int(h*C*Theta(x), x=0..2*L);**

`> `
**simplify(convert(eqn,trig));**

`> `
**a:=convert(exp(m*L)-exp(-m*L),trig):**

`> `
**b:=simplify(convert (exp(m*L)+exp(-m*L),trig)):**

`> `
**simplify(a/b);**

`> `
**eqn:=2*q[0]=h*C*Theta0*sinh(m*L)/(m*cosh(m*L));**

Heat Conduction in the Steady State