Physics - Heat & Thermo Question with Solution | TestHub

In steady state, the rate of heat flow $Q$ is constant along the rod. Fourier's law for conduction is $Q=-k(T)A\frac{dT}{dx}$. Rearranging and integrating from $x=0$ to $x=L$ and $T=T_1$ to $T=T_2$:$$Q{\int }_0^{L}dx=-A{\int }_{T_1}^{T_2}k(T)dT$$ $$QL=-A{\int }_{T_1}^{T_2}(0.01)T^{2}dT=-0.01A{\left[\frac{T^3}{3}\right]}_{T_1}^{T_2}=-0.01A\left(\frac{T_2^3-T_1^3}{3}\right)$$ Solving for $Q$: $Q=\frac{0.01A(T_1^3-T_2^3)}{3L}$. Substituting the given values: $A=10^{-4}{\text{m}}^2$, $L=1\text{m}$, $T_1=300\text{K}$, $T_2=400\text{K}$.$$Q=\frac{0.01\times 10^{-4}\times (300^3-400^3)}{3\times 1}=\frac{10^{-6}\times (27\times 10^6-64\times 10^6)}{3}$$ $$Q=\frac{10^{-6}\times (-37\times 10^6)}{3}=-\frac{37}{3}\approx -12.333\text{W}$$

The magnitude of the rate of heat flow is $12.333\text{W}$ (the negative sign indicates heat flows from $T_2$ to $T_1$).