Question

When $M_1$ gram of ice at -5° C (specific heat = 0.5 cal g⁻¹ °C⁻¹) is added to $M_2$ gram of water at 50°C (specific heat = 1 cal g⁻¹ °C⁻¹), finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g⁻¹ is:

• A. $\frac{5M_1}{M_2} - 50$
• B. $\frac{50M_2}{M_1}$
• C. $\frac{50M_2}{M_1} - 2.5$
• D. $\frac{50M_2}{M_1} - 5$

Answer 1

Answer: (C)

(3) $\frac{50M_2}{M_1} - 2.5$

Answer 2

The problem involves the principle of calorimetry, which states that in an isolated system, the total heat lost by hotter bodies is equal to the total heat gained by colder bodies. In this case, the water at $50^\circ C$ loses heat, and the ice at $-5^\circ C$ gains heat. The final state is water at $0^\circ C$, meaning all ice has melted.

1. Heat gained by ice ($Q_{gain}$):

The ice undergoes two processes to reach the final state:

• Heating of ice: Ice at $-5^\circ C$ heats up to $0^\circ C$.

Heat gained, $Q_1 = M_1 \times s_{ice} \times \Delta T_{ice}$

$Q_1 = M_1 \times 0.5 \text{ cal g}^{-1} {}^\circ C^{-1} \times (0 - (-5))^\circ C$

$Q_1 = M_1 \times 0.5 \times 5 = 2.5 M_1 \text{ cal}$

• Melting of ice: Ice at $0^\circ C$ melts into water at $0^\circ C$.

Heat gained, $Q_2 = M_1 \times L_f$

where $L_f$ is the latent heat of fusion of ice.

Total heat gained by ice, $Q_{gain} = Q_1 + Q_2 = 2.5 M_1 + M_1 L_f$.

2. Heat lost by water ($Q_{lost}$):

The water at $50^\circ C$ cools down to $0^\circ C$.

Heat lost, $Q_{lost} = M_2 \times s_{water} \times \Delta T_{water}$

$Q_{lost} = M_2 \times 1 \text{ cal g}^{-1} {}^\circ C^{-1} \times (50 - 0)^\circ C$

$Q_{lost} = M_2 \times 1 \times 50 = 50 M_2 \text{ cal}$

3. Applying the principle of calorimetry:

Heat Lost = Heat Gained

$Q_{lost} = Q_{gain}$

$50 M_2 = 2.5 M_1 + M_1 L_f$

4. Solving for $L_f$:

Rearrange the equation to find $L_f$:

$M_1 L_f = 50 M_2 - 2.5 M_1$

$L_f = \frac{50 M_2 - 2.5 M_1}{M_1}$

$L_f = \frac{50 M_2}{M_1} - \frac{2.5 M_1}{M_1}$

$L_f = \frac{50 M_2}{M_1} - 2.5 \text{ cal g}^{-1}$