Question

1 kg of ice at – 10 ° C is mixed with 4 . 4 kg of water at 30 ° C . The final temperature of mixture is

• A. 235/27 °C
• B. 10 °C
• C. 0 °C
• D. 30 °C

Answer 1

Answer: (A)

235/27 °C

Answer 2

1. Heat to warm ice to 0°C ($Q_1$): $Q_1 = m_{ice} \cdot s_{ice} \cdot \Delta T = 1000 \text{ g} \cdot 0.5 \text{ cal/g°C} \cdot (0 - (-10)^\circ\text{C}) = 5000$ cal.

2. Heat to melt ice at 0°C ($Q_2$): $Q_2 = m_{ice} \cdot L_f = 1000 \text{ g} \cdot 80 \text{ cal/g} = 80000$ cal. Total heat to convert ice to water at 0°C = $Q_1 + Q_2 = 85000$ cal.

3. Heat available from water cooling to 0°C ($Q_{water\_to\_0}$): $Q_{water\_to\_0} = m_{water} \cdot s_{water} \cdot \Delta T = 4400 \text{ g} \cdot 1 \text{ cal/g°C} \cdot (30^\circ\text{C} - 0^\circ\text{C}) = 132000$ cal.

4. Since $132000$ cal $> 85000$ cal, all ice melts, and the final temperature will be above 0°C.

5. Heat balance for equilibrium temperature $T$: Heat lost by water = Heat gained by ice (to melt) + Heat gained by melted ice (to reach final temperature $T$) $m_{water} \cdot s_{water} \cdot (30^\circ\text{C} - T) = (Q_1 + Q_2) + m_{ice} \cdot s_{water} \cdot (T - 0^\circ\text{C})$ $4400 \cdot 1 \cdot (30 - T) = 85000 + 1000 \cdot 1 \cdot T$ $132000 - 4400T = 85000 + 1000T$ $47000 = 5400T$ $T = \frac{47000}{5400} = \frac{470}{54} = \frac{235}{27}^\circ\text{C}$