Question

A spherical shell of radius 1 m, made of a steel sheet of thickness 5 mm and completely filled with ice at $0^\circ$C is immersed in boiling water. The time taken for the ice to melt completely is (Thermal conductivity of steel = $45~\text{Wm}^{-1}\text{K}^{-1}$ and density of ice is $0.9~\text{g cm}^{-3}$)

• A. 224 s
• B. 112 s
• C. 186 s
• D. 56 s

Hint:

Remember: \textbf{Rate of heat conduction} through a material: $Q = \frac{K A \Delta T}{d}$. Always check units of $K$, area, thickness, and $\Delta T$.
\textbf{Stepwise approach:} 1. Calculate area of the shell. 2. Convert thickness to meters. 3. Calculate $Q$ per second. 4. Calculate total heat needed to melt ice using $m L$. 5. Divide total heat by heat per second to get time.
Tip: For spheres, $A = 4 \pi R^2$ and $V = \frac{4}{3}\pi R^3$. Always convert grams to kg and cm to meters.

Answer 1

The Correct Option is B

1. Heat conducted through the steel shell per unit time: $Q = \frac{K A \Delta T}{d}$, where $K$ = thermal conductivity, $A$ = area, $\Delta T$ = temperature difference, $d$ = thickness.
2. Surface area of sphere: $A = 4 \pi R^2 = 4 \pi (1)^2 = 12.566~\text{m}^2$
3. Thickness of steel: $d = 5~\text{mm} = 0.005~\text{m}$
4. $\Delta T = 100 - 0 = 100~\text{K}$
5. Heat conducted per second: $Q = \frac{45 \cdot 12.566 \cdot 100}{0.005} \approx 1.131 \times 10^6~\text{W}$
6. Mass of ice: $V = \frac{4}{3}\pi R^3 = 4.188~\text{m}^3$, $\rho = 900~\text{kg/m}^3 \implies m = 3769~\text{kg}$ (Check units: $0.9~\text{g/cm}^3 = 900~\text{kg/m}^3$)
7. Heat required to melt ice: $Q_\text{ice} = m L = 3769 \cdot 3.36 \times 10^5 \approx 1.27 \times 10^9~\text{J}$
8. Time required: $t = \frac{Q_\text{ice}}{Q} \approx \frac{1.27 \times 10^9}{1.131 \times 10^6} \approx 112~\text{s}$