A 0.5970-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C.

Question

A 0.5970-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 6.550 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

The source below gives all the necessary constants.
The question implies that the final state will be liquid. So first turn the two states into liquids.

(2.087 J/(g·°C) x (575.0 g ice) x (0 - (-12.40))°C = 14880 J to warm the ice to 0°C
(333.6 J/g) x (575.0 g ice) = 191820 J to melt the ice

(36.03 J/(mol·°C)) x (6.550 mol steam) x (365.0°C - 100°C) = 62539 J to cool the steam to 100°C
(40660 J/mol) x (6.550 mol steam) = 266323 J to condense the steam

So far the steam has lost 62539 J + 266323 J = 328862 J
and the ice has gained 14880 J + 191820 J = 206700 J
The steam provided the excess amount of heat: 328862 J - 206700 J = 122162 J
Put this excess heat into the cooler water:
(122162 J) / (4.184 J/(g·°C) / (575.0 g) = 50.778 °C change

So now we have 6.550 mol of liquid water at 100°C and 575.0 g of liquid water at 50.778 °C.
Change the mol to g: (6.550 mol H2O) x (18.01532 g H2O/mol) = 118.00 g
Figure the weighted average:
((100°C x 575.0 g ) + (50.778°C x 118.00 g)) / (575.0 g + 118.00 g) = 91.62°C

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