Question

A metal rod of cross sectional area 1.0 cm2 is being heated at one end. At one time, the temperatures gradient is 5.0°C cm−1 at cross section A and is 2.5°C cm−1 at cross section B. Calculate the rate at which the temperature is increasing in the part AB of the rod. The heat capacity of the part AB = 0.40 J°C−1, thermal conductivity of the material of the rod = 200 W m−1°C−1. Neglect any loss of heat to the atmosphere

Answer 1

The rate at which the temperature is increasing in the part AB of the rod is $\frac{d T}{d t}=12.5^{\circ} \mathrm{C} / \mathrm{sec}$

Explanation:

Step 1:

Given,      

Area of cross section of metal rod = $1 \mathrm{cm}^{2}$

Temperature gradient at cross section $A=5.0^{\circ} \mathrm{Ccm}^{-1}$

Temperature gradient at cross section $B=2.5^{\circ} \mathrm{Ccm}^{-1}$

Heat capacity of AB part = $0.40 \mathrm{J}^{\circ} \mathrm{C}^{-1}$

Thermal conductivity of material of rod = $200 \mathrm{Wm}^{-1 \circ} \mathrm{C}^{-1}$

Rate of heat flow at A end is $\frac{\Delta Q_{A}}{\Delta t}=\frac{K A d\left(T_{A}\right)}{d l}$

Step 2:

Similarly rate of heat flow at B end is  $\frac{\Delta Q_{B}}{\Delta t}=\frac{K A d\left(T_{B}\right)}{d l}$

Heat absorbed $\mathrm{dQ}=\mathrm{ms} \Delta T$

Step 3:

Rate of heat absorbed by rod is $\frac{d Q}{d t}=m s \frac{d T}{d t}$

$m s \frac{d T}{d t}=\frac{K A d\left(T_{A}\right)}{d l}-\frac{K A d\left(T_{B}\right)}{d l}$

$0.4 \frac{d T}{d t}=200 * 10^{-4}[5-2.5]$

$\frac{d T}{d t}=200 * 2.5 * 10^{-4} / 0.4$

$\frac{d T}{d t}=12.5^{0} \mathrm{C} / \mathrm{sec}$