One end of a metal rod having a cross-sectional area of 1.0 cm^2 is...

The rate of heat flow at cross-section $A$ is given by Fourier's law of heat conduction:

$H_A = K A \left| \dfrac{dT}{dx} \right|_A$

Substituting the given values:

$H_A = 200 \times (1.0 \times 10^{-4}) \times (5.0 \times 10^2) = 10 \text{ W}$

Similarly, the rate of heat flow at cross-section $B$ is:

$H_B = K A \left| \dfrac{dT}{dx} \right|_B$

$H_B = 200 \times (1.0 \times 10^{-4}) \times (2.5 \times 10^2) = 5 \text{ W}$

The net rate of heat accumulation in the section $AB$ is the difference between the heat entering at $A$ and leaving at $B$:

$\dfrac{dQ}{dt} = H_A - H_B = 10 - 5 = 5 \text{ W}$

The rate at which the temperature is increasing in section $AB$ is related to the net heat accumulation by:

$\dfrac{dT}{dt} = \dfrac{1}{C} \dfrac{dQ}{dt}$

Given the heat capacity C = 0.40 \text{ J }^{}^{{}^{\circ}}\textC^{-1}:

\dfrac{dT}{dt} = \dfrac{5}{0.40} = 12.5^{}^{{}^{\circ}}\text{C s}^{-1}