Question

Ice of mass \(80\,\mathrm{g}\) at a temperature of \(-10^\circ\mathrm{C}\) is mixed with water of mass \(100\,\mathrm{g}\) at a temperature of \(20^\circ\mathrm{C}\).

The ratio of the masses of ice and water in the mixture at equilibrium is 

Given:

• Latent heat of fusion of ice \(= 80\,\mathrm{cal\,g^{-1}}\)
• Specific heat capacity of water \(= 1\mathrm{cal\,g^{-1}^\circ C^{-1}}\)
• Specific heat capacity of ice \(= 0.5\mathrm{cal\,g^{-1}^\circ C^{-1}}\)

• A. \(4:5\)
• B. \(1:2\)
• C. \(2:3\)
• D. \(3:4\)

Hint:

When ice and water are mixed, \[ \boxed{ \text{Heat lost} = \text{Heat gained}. } \] Always account for: \[ \boxed{ \begin{aligned} &\text{Heating of ice}, &\text{Melting of ice}, &\text{Heating of melted water (if any).} \end{aligned} } \]

Answer 1

The Correct Option is B

Step 1: Calculate the heat released by water. Water cools from \[ 20^\circ\text{C} \] to \[ 0^\circ\text{C}. \] Hence, \[ Q_w = 100\times1\times20 = 2000\text{ cal}. \]

Step 2: Calculate the heat required by the ice. Heat required to raise the temperature of ice from \[ -10^\circ\text{C} \] to \[ 0^\circ\text{C} \] is \[ Q_1 = 80\times0.5\times10 = 400\text{ cal}. \] Remaining heat available for melting is \[ 2000-400 = 1600\text{ cal}. \] Mass of ice melted is \[ m = \frac{1600}{80} = 20\text{ g}. \]

Step 3: Find the masses at equilibrium. Ice remaining \[ = 80-20 = 60\text{ g}. \] Water present \[ = 100+20 = 120\text{ g}. \] Therefore, \[ \text{Ice}:\text{Water} = 60:120 = 1:2. \] Hence, \[ \boxed{1:2}. \] Thus, \[ \boxed{(B)} \] is the correct answer.