60 g of ice at 0°C is mixed with 60 g steam at 100°C. At thermal equilibrium, the mixture contains:

• A.) 80 g of water and 40 g of steam at 100°C
• B.) 120 g of water at 90°C
• C.) 120 g of water at 100°C
• D.) 40 g of steam and 80 g of water at 0°C

To solve this problem, we need to consider the heat exchange between the ice and the steam until they reach thermal equilibrium. The ice will absorb heat to melt into water, while the steam will release heat to condense into water.

Calculating Heat Transfer

First, let's calculate the heat required to melt the ice:

• Mass of ice = 60 g
• Heat of fusion of ice = 334 J/g
• Heat required to melt ice = 60 g × 334 J/g = 20,040 J

Next, we calculate the heat released by the steam when it condenses:

• Mass of steam = 60 g
• Heat of vaporization of steam = 2,260 J/g
• Heat released by steam = 60 g × 2,260 J/g = 135,600 J

Heat Balance

When the ice melts, it absorbs 20,040 J. The remaining heat from the steam after condensation is:

• Heat remaining = 135,600 J - 20,040 J = 115,560 J

This remaining heat will warm the resulting water from the melted ice. The total mass of water after melting the ice is:

• Total mass of water = 60 g (from ice) + 60 g (from steam) = 120 g

Final Temperature Calculation

Now, we can find the final temperature of the water:

• Specific heat of water = 4.18 J/g°C
• Using the formula: Q = mcΔT, we can rearrange it to find ΔT:

ΔT = Q / (m × c) = 115,560 J / (120 g × 4.18 J/g°C) ≈ 231.5°C

Since this temperature exceeds 100°C, all the steam will condense, and the final mixture will be:

• 120 g of water at 100°C

Final Answer

The correct option is:

C.) 120 g of water at 100°C