Question

5 g of ice at −10 ∘ C is mixed with 20 g of water at 30 DEGREE CENTIGRSDE ∘ C. The equilibrium temperature of mixture will be (specific heat of ice=0.5 g ∘ C Cal ​ )

• A. 10 ∘ C
• B. 5 ∘ C
• C. 7 ∘ C
• D. 0 ∘ C

Answer 1

Answer: (C)

7 ∘ C

Answer 2

1. Heat to warm ice to 0°C ($Q_1$): $Q_1 = m_{ice} \cdot s_{ice} \cdot \Delta T = 5 \text{ g} \cdot 0.5 \text{ Cal/g°C} \cdot (0 - (-10)^\circ\text{C}) = 25$ Cal.

2. Heat to melt ice at 0°C ($Q_2$): $Q_2 = m_{ice} \cdot L_f = 5 \text{ g} \cdot 80 \text{ Cal/g} = 400$ Cal. Total heat to convert ice to water at 0°C = $Q_1 + Q_2 = 425$ Cal.

3. Heat available from water cooling to 0°C ($Q_{water\_to\_0}$): $Q_{water\_to\_0} = m_{water} \cdot s_{water} \cdot \Delta T = 20 \text{ g} \cdot 1 \text{ Cal/g°C} \cdot (30^\circ\text{C} - 0^\circ\text{C}) = 600$ Cal.

4. Since $600$ Cal $> 425$ Cal, all ice melts, and the final temperature will be above 0°C.

5. Heat balance for equilibrium temperature $T$: Heat gained by ice (warming from 0°C to $T$) = Heat lost by water (cooling from 30°C to $T$). $(Q_1 + Q_2) + m_{ice} \cdot s_{water} \cdot (T - 0^\circ\text{C}) = m_{water} \cdot s_{water} \cdot (30^\circ\text{C} - T)$ $425 + 5 \cdot 1 \cdot T = 20 \cdot 1 \cdot (30 - T)$ $425 + 5T = 600 - 20T$ $25T = 175$ $T = 7^\circ$C