Question

The length of a metal rod is 20 cm and its area of cross-section is \( 4 \, \text{cm}^2 \). If one end of the rod is kept at a temperature of \( 100^\circ\text{C} \) and the other end is kept in ice at \( 0^\circ\text{C} \), then the mass of the ice melted in 7 minutes is (Thermal conductivity of the metal = \( 90 \, \text{Wm}^{-1}\text{K}^{-1} \) and latent heat of fusion of ice = \( 336 \times 10^3 \, \text{Jkg}^{-1} \))

• A. 90 g
• B. 67.5 g
• C. 22.5 g
• D. 45 g

Hint:

Ensure all units are converted to SI units (meters, seconds, Joules, kg) before calculation to avoid errors with powers of 10.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

The heat conducted through the metal rod is used to melt the ice. We need to equate the rate of heat flow through the rod to the rate of heat absorption by the ice for fusion.
Step 2: Key Formula or Approach:

1. Rate of heat flow: \( \frac{Q}{t} = \frac{KA(T_1 - T_2)}{l} \) 2. Heat for phase change: \( Q = mL_f \)
Step 3: Detailed Explanation:

Given: Length \( l = 20 \, \text{cm} = 0.2 \, \text{m} \) Area \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \) Temperature difference \( \Delta T = 100 - 0 = 100 \, \text{K} \) Time \( t = 7 \, \text{minutes} = 7 \times 60 = 420 \, \text{s} \) Thermal Conductivity \( K = 90 \, \text{Wm}^{-1}\text{K}^{-1} \) Latent Heat \( L_f = 336 \times 10^3 \, \text{Jkg}^{-1} \) First, calculate the rate of heat flow: \[ \frac{dQ}{dt} = \frac{90 \times (4 \times 10^{-4}) \times 100}{0.2} \] \[ \frac{dQ}{dt} = \frac{360 \times 10^{-2}}{0.2} = \frac{3.6}{0.2} = 18 \, \text{J/s} \] Total heat transferred in 420 seconds: \[ Q = 18 \times 420 = 7560 \, \text{J} \] Now, find the mass of ice melted using \( Q = mL_f \): \[ m = \frac{Q}{L_f} = \frac{7560}{336 \times 10^3} \, \text{kg} \] \[ m = \frac{7560}{336000} \, \text{kg} = 0.0225 \, \text{kg} \] Converting to grams: \[ m = 0.0225 \times 1000 \, \text{g} = 22.5 \, \text{g} \]
Step 4: Final Answer:

The mass of ice melted is 22.5 g.