Chapter 28, Problem 43P

The rate of heat flow (conduction) between two points on a cylinder heated at one end is given by

$\frac{dQ}{dt}=\lambda A\frac{dT}{dx}$

Where $\lambda =$ a constant, $A=$ the cylinder's cross-sectional area, $Q=$ heat flow, $T=$ temperature, $t=$ time, and $x=$ distance from the heated end. Because the equation involves two derivatives, wewill simplify this equation by letting

$\frac{dT}{dx}=\frac{100\left(L-x\right)\left(20-t\right)}{100-xt}$

where L is the length of the rod. Combine the two equations and compute the heat flow for $t=0$ to 25 s. The initial condition is $Q\left(0\right)=0$ and the parameters are, $A=12{\text{ cm}}^2,L=20\text{ cm}$, and $x=2.5$ cm. Plot your results.