Question

A uniform copper rod of 50 cm length is insulated on the sides, and has its ends exposed to ice and steam respectively. If there is a layer of water 1 mm thick at each end, the temperature gradient (in $^\circ\text{C m}^{-1}$) in the bar is (Assume that the thermal conductivity of copper is 400 $\text{W m}^{-1}\text{K}^{-1}$ and water is 0.4 $\text{W m}^{-1}\text{K}^{-1}$)

• A. 60
• B. 40
• C. 50
• D. 55
• E. 65

Hint:

The temperature gradient is inversely proportional to the thermal conductivity. Because copper is a great conductor, its gradient is much lower than that of the water layers.

Answer 1

The Correct Option is B

Concept:
The system acts like three thermal resistors in series: water layer, copper rod, and water layer. In a series combination, the rate of heat flow ($Q/t$) is the same through all layers. \[ \frac{Q}{t} = \frac{\Delta T_{total}}{R_{total}} \] Thermal Resistance $R = \frac{L}{KA}$.

Step 1: Calculate the total thermal resistance.
Let $A$ be the area. $R_{water} = \frac{10^{-3}}{0.4 A} = \frac{0.0025}{A}$ per layer. Total $R_{water} = \frac{0.005}{A}$. $R_{copper} = \frac{0.5}{400 A} = \frac{0.00125}{A}$. $R_{total} = \frac{0.005 + 0.00125}{A} = \frac{0.00625}{A}$.

Step 2: Find the heat flow rate.
Total temperature difference $\Delta T = 100 - 0 = 100^\circ$C. \[ \frac{Q}{At} = \frac{100}{0.00625} = 16000 \text{ W/m}^2 \]

Step 3: Determine the temperature gradient in the copper rod.
Temperature gradient in copper $= \frac{1}{K_{cu}} \left(\frac{Q}{At}\right)$ \[ \text{Gradient} = \frac{16000}{400} = 40^\circ\text{C m}^{-1} \]