Question

The length of a metal bar is 20 cm and the area of cross-section is \( 4 \times 10^{-4} \) m\(^2\). If one end of the rod is kept in ice at \( 0^\circ C \) and the other end is kept in steam at \( 100^\circ C \), the mass of ice melted in one minute is 5 g. The thermal conductivity of the metal in Wm\(^{-1}\)K\(^{-1}\) is (Latent heat of fusion = 80 cal/gm)

• A. 140
• B. 120
• C. 100
• D. 160

Hint:

In heat transfer problems, always convert all units to SI before calculations.

Answer 1

The Correct Option is A

The heat transfer equation is given by Fourier’s Law: \[ Q = \frac{k A \Delta T}{L} t \] Heat required to melt ice: \[ Q = m L \] Given: \[ m = 5g = 5 \times 10^{-3} kg, \quad L = 80 \text{ cal/g} = 80 \times 4.18 \text{ J/g} \] \[ Q = 5 \times 10^{-3} \times 80 \times 4.18 \] \[ = 1.672 \text{ kJ} = 1672 \text{ J} \] Now, using Fourier’s equation: \[ 1672 = \frac{k \times 4 \times 10^{-4} \times 100}{0.2} \times 60 \] Solving for \( k \), we get: \[ k = 140 \text{ Wm\(^{-1}\)K\(^{-1}\)} \] Thus, the correct answer is 140.