A 0.6410-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C.

Question

A 0.6410-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 5.830 moles of steam (water) molecules before the ice is added, calculate the final temperature of the puddle once it settled to equilibrium. (Assume the chamber walls are sufficiently flexible to allow the system to remain isobaric and consider thermal losses/gains from the chamber walls as negligible.) Use the following values for the heat capacities of ice, water, and steam. ice= 2093 J/(kg*C) water= 4186 J/(kg*C) steam= 2009 J/(kg*C)

Explanation / Answer

molecular weight of water (about 18 grams, obviously) -
so the mass of steam is (5.83 mol)(0.018 kg/mol) = 0.105 kg.
Now,
Heat capacity of ice is 2093 J/(kg K).

The heat needed to warm up the ice to 0C is (2.093J)(0.641)(12.4) = 15.94 kJ
The heat released as the steam cools down to 100C is (2.009 kJ)(0.105)(265) = 55.9 kJ
The heat needed to melt the ice at 0C is (334 kJ)(0.641) = 210 kJ
The heat released as the steam condenses at 100C is (2.009 kJ)(0.105) = 0.211 kJ
The heat needed to warm the melted ice to the equilibrium T of the puddle is
(4.18 kJ)(0.641)(T in deg C)
The heat released as the condensed steam cools to equilibrium T is
(4.18 kJ)(0.105)(100 - T)
So,
15.94kJ + 210kJ + (4.18 kJ)(0.641) T = 55.9kJ + 0.211kJ + (4.18 kJ)(0.120) (100 - T)
Then solve for T.

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