Question

A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100$^{\circ}$C. A block of ice at 0$^{\circ}$C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice = $ 3.36 \times 10^5 J kg^{-1})$

• A. $ 1.24J/m/s / \circ C$
• B. $ 1.29 J/m/s / \circ C$
• C. $ 2.05 J/m/s / \circ C$
• D. $ 1.02 J/m/s / \circ C$

Answer 1

The Correct Option is A

To solve this problem, we need to determine the thermal conductivity of the slab of stone that allows heat transfer from the steam to the ice, thus melting a certain amount of ice. We will use the formula for heat conduction and the latent heat of fusion of ice.

1. First, calculate the amount of heat required to melt 4.8 kg of ice.
   • Given:
     Latent heat of fusion of ice, L = 3.36 \times 10^5 \, \text{J kg}^{-1}
     Mass of ice melted, m = 4.8 \, \text{kg}
   • Heat required, Q = m \times L
   • So, Q = 4.8 \times 3.36 \times 10^5 = 1.6128 \times 10^6 \, \text{J}
2. Since the heat is transferred over the period of one hour, convert time:
   • Time, t = 1 \, \text{hour} = 3600 \, \text{seconds}
3. Apply the formula for heat transfer by conduction through the slab:
   • Heat transferred, Q = \frac{{K \cdot A \cdot \Delta T \cdot t}}{{d}}
   • Where:
     • K is the thermal conductivity of the slab.
     • A = 0.36 \, \text{m}^2 is the area of the slab's surface.
     • \Delta T = 100 - 0 = 100^{\circ}C is the temperature difference across the slab.
     • d = 0.1 \, \text{m} is the thickness of the slab.
     • t = 3600 \, \text{s} is the time period.
   • Rearrange to solve for K:
   • K = \frac{{Q \cdot d}}{{A \cdot \Delta T \cdot t}}
   • Substitute the known values:
   • K = \frac{{1.6128 \times 10^6 \cdot 0.1}}{{0.36 \cdot 100 \cdot 3600}}
   • Calculate K = 1.24 \, \text{J/m/s/}^{\circ}\text{C}

Thus, the thermal conductivity of the slab is 1.24 J/m/s/°C, which matches the correct option.