HC Verma Chapter 25 Problem 39

Problem Statement

200 g of ice at 0°C is added to 500 g of water at 40°C. Find the final temperature. L_f = 3.36×10⁵ J/kg, c_w = 4200 J/(kg·K).

Given Information

• m_ice = 200 g = 0.2 kg, T_ice = 0°C
• m_w = 500 g = 0.5 kg, T_w = 40°C
• L_f = 3.36×10⁵ J/kg
• c_w = 4200 J/(kg·K)

Physical Concepts & Formulas

First check if the water has enough energy to melt all the ice. Heat available from water cooling to 0°C versus heat needed to melt all ice.

• Heat available: $Q_{avail} = m_w c_w \Delta T$
• Heat to melt all ice: $Q_{melt} = m_{ice} L_f$
• Compare and apply calorimetry

Step-by-Step Solution

1. Heat available from water cooling to 0°C: $Q_{avail} = 0.5 \times 4200 \times 40 = 84000$ J
2. Heat to melt all ice: $Q_{melt} = 0.2 \times 3.36 \times 10^5 = 67200$ J
3. Since $Q_{avail} > Q_{melt}$, all ice melts; remaining heat = $84000 – 67200 = 16800$ J
4. Total water mass after melting = $0.5 + 0.2 = 0.7$ kg
5. Final temp: $T_f = \dfrac{Q_{remaining}}{m_{total} c_w} = \dfrac{16800}{0.7 \times 4200} = \dfrac{16800}{2940} \approx 5.7°C$

Worked Calculation

$T_f = \dfrac{84000 – 67200}{0.7 \times 4200} = \dfrac{16800}{2940} \approx 5.7°C$

Answer

T_f ≈ 5.7°C

Physical Interpretation

All ice melts and the final temperature is about 5.7°C — well above 0°C. The water had enough energy to melt all ice with heat left over. If ice were more (say 300 g), not all would melt and the final temperature would be 0°C.