Question

A 100 cm long cylindrical flask with inner and outer diameters 2 cm and 4 cm respectively is completely filled with ice at 0°C as shown in the figure. The constant temperature outside the flask is 40°C, then :- (Thermal conductivity of the flask is 0.693 W/m°C, L$_{ice}$ = 80 cal/gm)

• A. Rate of heat flow from outside to the flask is 80$\pi$ J/s.
• B. The rate at which ice melts is $\frac{\pi}{4200}$ Kg/s.
• C. The rate at which ice melts is 100$\pi$ Kg/s.
• D. Rate of heat flow from outside to flask is 40$\pi$ J/s.

Answer 1

Answer: (A), (B)

The correct options are (A) and (B).

Answer 2

1. Heat Flow Calculation: The rate of heat flow ($H$) through a cylindrical wall is given by: $H = \frac{2 \pi K L (T_{outer} - T_{inner})}{\ln(r_{outer}/r_{inner})}$ Given values:

   • Length, $L = 100 \, \text{cm} = 1 \, \text{m}$
   • Inner radius, $r_1 = \frac{2 \, \text{cm}}{2} = 1 \, \text{cm} = 0.01 \, \text{m}$
   • Outer radius, $r_2 = \frac{4 \, \text{cm}}{2} = 2 \, \text{cm} = 0.02 \, \text{m}$
   • Thermal conductivity, $K = 0.693 \, \text{W/m}^\circ\text{C}$
   • Temperature difference, $\Delta T = T_{outer} - T_{inner} = 40^\circ\text{C} - 0^\circ\text{C} = 40^\circ\text{C}$

   Substituting these values: $H = \frac{2 \pi (0.693 \, \text{W/m}^\circ\text{C}) (1 \, \text{m}) (40^\circ\text{C})}{\ln(0.02 \, \text{m} / 0.01 \, \text{m})}$ $H = \frac{2 \pi (0.693) (40)}{\ln(2)}$ Using the approximation $\ln(2) \approx 0.693$: $H = \frac{2 \pi (0.693) (40)}{0.693} = 2 \pi (40) = 80 \pi \, \text{J/s}$ This matches option (A).

2. Ice Melting Rate Calculation: The heat flowing into the ice is used for melting. The rate of melting ($\frac{dm}{dt}$) is given by: $\frac{dm}{dt} = \frac{H}{L_{ice}}$ Given latent heat of fusion, $L_{ice} = 80 \, \text{cal/gm}$. To use this in SI units, we convert calories to Joules and grams to kilograms. A common approximation used in such problems is $1 \, \text{cal} = 4.2 \, \text{J}$. Therefore, $1 \, \text{cal/gm} = 4.2 \, \text{J/gm} = 4.2 \times 1000 \, \text{J/kg} = 4200 \, \text{J/kg}$. So, $L_{ice} = 80 \times 4200 \, \text{J/kg} = 336000 \, \text{J/kg}$.

   Now, calculate the rate of melting: $\frac{dm}{dt} = \frac{80 \pi \, \text{J/s}}{336000 \, \text{J/kg}}$ $\frac{dm}{dt} = \frac{80 \pi}{336000} \, \text{kg/s}$ $\frac{dm}{dt} = \frac{\pi}{4200} \, \text{kg/s}$ This matches option (B).