Question

In the figure shown, AB is a rod of length 30 cm and area of cross-section 1.0 $cm^2$ and thermal conductivity 336 S.I. units. The ends A & B are maintained at temperatures 20° C and 40° C respectively. A point C of this rod is connected to a box D, containing ice at 0° C, through a highly conducting wire of negligible heat capacity. The rate at which ice melts in the box is : [Latent heat of fusion for ice L, = 80 cal/gm]

• A. 84 mg/s
• B. 84 g/s
• C. 20 mg/s
• D. 40 mg/s

Answer 1

Answer: (D)

40 mg/s

Answer 2

The problem involves heat conduction through a rod and the melting of ice. We are given a rod AB with ends maintained at temperatures $T_A = 20^\circ C$ and $T_B = 40^\circ C$. Point C on the rod is connected to ice at $T_C = 0^\circ C$ via a highly conducting wire, implying that the temperature at point C is $T_C = 0^\circ C$.

The rod has length $L_{AB} = 30$ cm, area of cross-section $A = 1.0 \, cm^2$, and thermal conductivity $k = 336$ S.I. units. Point C is located 10 cm from A ($L_{AC} = 10$ cm) and 20 cm from B ($L_{CB} = 20$ cm). The latent heat of fusion for ice is $L_f = 80$ cal/gm.

Since $T_A > T_C$, heat flows from end A to point C. The rate of heat flow through section AC is given by Fourier's law: $H_{AC} = k A \frac{T_A - T_C}{L_{AC}}$

Since $T_B > T_C$, heat flows from end B to point C. The rate of heat flow through section CB is: $H_{CB} = k A \frac{T_B - T_C}{L_{CB}}$

We need to convert all units to SI units for calculation. $A = 1.0 \, cm^2 = 1.0 \times 10^{-4} \, m^2$ $L_{AC} = 10 \, cm = 0.1 \, m$ $L_{CB} = 20 \, cm = 0.2 \, m$ $k = 336 \, W/(m \cdot K)$

Now, calculate the heat flow rates: $H_{AC} = 336 \, W/(m \cdot K) \times (1.0 \times 10^{-4} \, m^2) \times \frac{20^\circ C - 0^\circ C}{0.1 \, m} = 336 \times 10^{-4} \times \frac{20}{0.1} = 336 \times 10^{-4} \times 200 = 6.72 \, W$

$H_{CB} = 336 \, W/(m \cdot K) \times (1.0 \times 10^{-4} \, m^2) \times \frac{40^\circ C - 0^\circ C}{0.2 \, m} = 336 \times 10^{-4} \times \frac{40}{0.2} = 336 \times 10^{-4} \times 200 = 6.72 \, W$

The total rate at which heat is transferred to point C from the rod is the sum of these two heat flows: $H_{melt} = H_{AC} + H_{CB} = 6.72 \, W + 6.72 \, W = 13.44 \, W$

This heat is used to melt the ice. The rate of melting of ice ($\frac{dm}{dt}$) is given by $H_{melt} = \frac{dm}{dt} \times L_f$. We need to use consistent units for $L_f$. Given $L_f = 80$ cal/gm. We use the conversion factor 1 cal = 4.184 J. $L_f = 80 \frac{\text{cal}}{\text{gm}} = 80 \times \frac{4.184 \, J}{10^{-3} \, kg} = 80 \times 4184 \, J/kg = 334720 \, J/kg$.

Now, calculate the rate of mass melting: $\frac{dm}{dt} = \frac{H_{melt}}{L_f} = \frac{13.44 \, J/s}{334720 \, J/kg} \approx 4.015296 \times 10^{-5} \, kg/s$

The options are in mg/s. Convert kg/s to mg/s: $1 \, kg = 10^6 \, mg$. $\frac{dm}{dt} \approx 4.015296 \times 10^{-5} \, kg/s \times 10^6 \, mg/kg = 40.15296 \, mg/s$.

Rounding to the nearest option, the rate at which ice melts is approximately 40 mg/s.