A 0.6630-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C.

Question

A 0.6630-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 6.310 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.)

Explanation / Answer

heat capacities of ice, steam, and water (no, they're not all the same);
latent heats of fusion and condensation (334 kJ/kg and 2260 kJ/kg);
molecular weight of water (about 18 grams, obviously) -
so the mass of steam is (6.31 mol)(0.018 kg/mol) = 0.1135 kg.

Various references give isobaric heat capacity of steam as:
2.08 kJ/(kg K) at 100C, 2.048 kJ/(kg K) at 365 C - so I'm going to
use 2.064 kJ/(kg K) as an average value for the heat capacity of the steam.
Heat capacity of water varies between 0 and 100C, but I'm going to use 4.18 kJ/(kg K).
Heat capacity of ice at not too deep a freeze is about 2.11 kJ/(kg K).

The heat needed to warm up the ice to 0C is (2.11 kJ)(0.663)(12.4)
= 17.3 kJ,
The heat released as the steam cools down to 100C is (2.064 kJ)(0.1135)(265)
= 62 kJ
The heat needed to melt the ice at 0C is (334 kJ)(0.663)
= 221.4 kJ
The heat released as the steam condenses at 100C is (2260 kJ)(0.1135)
= 256.5 kJ,.
The heat needed to warm the melted ice to the equilibrium T of the puddle is
(4.18 kJ)(0.663)(T in deg C)
The heat released as the condensed steam cools to equilibrium T is
(4.18 kJ)(0.1135)(100 - T)



17.3kJ + 221.4kJ + (4.18 kJ)(0.663) T = 62kJ + 256.5kJ + (4.18 kJ)(0.1135) (100 - T)
238.7+2.771 T=318.5+ 0.4744(100-T)

                        = 318.5+ 47.44-0.4744 T

127.27=3.2454 T

T= 39.215 C

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