Question

5 gm of ice at 0°C is mixed with 1gm steam at 100°C. The final temperature and composition of the mixture is

• A. Water at 60°C
• B. Water at 40°C
• C. 2 g water at 100°C and rest as steam
• D. 2 g water at 0°C and rest ice

Answer 1

Answer: (B)

Water at 40°C

Answer 2

Constants:

• Latent heat of fusion of ice ($L_f$) = 80 cal/gm
• Latent heat of vaporization of water ($L_v$) = 540 cal/gm
• Specific heat of water ($c_w$) = 1 cal/gm°C

Step 1: Calculate heat required to melt ice. Heat required to melt 5 gm of ice at 0°C into water at 0°C: $Q_{melt} = m_{ice} \times L_f = 5 \text{ gm} \times 80 \text{ cal/gm} = 400 \text{ cal}$

Step 2: Calculate heat released by steam to condense. Heat released by 1 gm of steam at 100°C to condense into water at 100°C: $Q_{condense} = m_{steam} \times L_v = 1 \text{ gm} \times 540 \text{ cal/gm} = 540 \text{ cal}$

Step 3: Determine if all ice melts. Since the heat released by condensation (540 cal) is greater than the heat required to melt all the ice (400 cal), all the ice will melt.

Step 4: Calculate the excess heat. After melting the ice, the remaining heat available from the steam is: $Q_{rem} = Q_{condense} - Q_{melt} = 540 \text{ cal} - 400 \text{ cal} = 140 \text{ cal}$ At this point, we have 5 gm of water at 0°C and 1 gm of water at 100°C. The 140 cal of remaining heat will be used to raise the temperature of this mixture.

Step 5: Calculate the final equilibrium temperature. Let the final equilibrium temperature be $T_f$. The total mass of water is $5 \text{ gm} + 1 \text{ gm} = 6 \text{ gm}$. The heat absorbed by the 5 gm of water to reach $T_f$ is $Q_{gain} = m_{w1} \times c_w \times (T_f - 0^\circ C) = 5 \times 1 \times T_f = 5T_f$. The heat released by the 1 gm of water to reach $T_f$ is $Q_{loss} = m_{w2} \times c_w \times (100^\circ C - T_f) = 1 \times 1 \times (100 - T_f) = 100 - T_f$.

Using conservation of energy, the heat lost by the hotter water must equal the heat gained by the colder water plus the excess heat. $Q_{loss} = Q_{gain} + Q_{rem}$ $100 - T_f = 5T_f + 140$ $100 - 140 = 5T_f + T_f$ $-40 = 6T_f$ $T_f = -40/6 \approx -6.67^\circ C$.

This result is not physically possible. A more robust method is to use the conservation of energy by calculating total heat absorbed and released, assuming a final state of liquid water.

Method 2: Using $Q_{absorbed} = Q_{released}$

Assume the final temperature $T_f$ is between 0°C and 100°C. The final mixture will be entirely water.

• Heat absorbed ($Q_{abs}$) by 5 gm ice to become water at $T_f$: $Q_{abs} = (m_{ice} \times L_f) + (m_{ice} \times c_w \times (T_f - 0^\circ C))$ $Q_{abs} = (5 \times 80) + (5 \times 1 \times T_f) = 400 + 5T_f$

• Heat released ($Q_{rel}$) by 1 gm steam to become water at $T_f$: $Q_{rel} = (m_{steam} \times L_v) + (m_{steam} \times c_w \times (100^\circ C - T_f))$ $Q_{rel} = (1 \times 540) + (1 \times 1 \times (100 - T_f)) = 540 + 100 - T_f = 640 - T_f$

• Equating heat absorbed and released: $Q_{abs} = Q_{rel}$ $400 + 5T_f = 640 - T_f$ $5T_f + T_f = 640 - 400$ $6T_f = 240$ $T_f = \frac{240}{6} = 40^\circ C$

Since the calculated temperature $T_f = 40^\circ C$ is between 0°C and 100°C, our assumption that the final state is liquid water at this temperature is correct. The final composition is the sum of the masses: 5 gm (from ice) + 1 gm (from steam) = 6 gm of water.