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 6.430 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.)

Answer #1

The data we need to do this problem are:

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 (6.43 mol)(0.018 kg/mol) = 0.120 kg.

Probably don't need "steam tables" - let's try it in a simpler
way.

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)(0.641)(12.4)

= about 15.62 kJ, but use calculator

The heat released as the steam cools down to 100C is (2.064
kJ)(0.12)(265)

= about 65 kJ, but use calculator.

The heat needed to melt the ice at 0C is (334 kJ)(0.597)

= about 199.40 kJ, but use calculator.

The heat released as the steam condenses at 100C is (2260
kJ)(0.12)

= about 270 kJ, but use calculator.

The heat needed to warm the melted ice to the equilibrium T of the
puddle is

(4.18 kJ)(0.597)(T in deg C)

The heat released as the condensed steam cools to equilibrium T
is

(4.18 kJ)(0.12)(100 - T)

When you have improved the accuracy of my 15.62, 65, 199.4, and
270 above,

you will have something looking roughly like this:

15.62kJ + 199.4kJ + (4.18 kJ)(0.597) T = 65kJ + 270kJ + (4.18
kJ)(0.120) (100 - T)

215KJ+2.50KJ*T=335KJ+0.5KJ(100-T)

2.50KJ * T-0.5KJ(100-T)=(335-215)=120KJ

3T=120+50=170

T=170/3=56.7C

final temperature of the puddle once it settled to equilibrium=57C

A 0.5750-kg. ice cube at -12.40 degree C is placed inside a
chamber of steam at 365.0 degree 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...

A 0.66300.6630 kg ice cube at −12.40−12.40 °C is placed inside a
rigid, thermally isolated chamber containing steam at 365.0365.0
°C. Later, you notice that the ice cube has completely melted into
a puddle of water. The specific heats of ice, water, and steam are
?ice=2093 J/(kg·∘C),cice=2093 J/(kg·∘C), ?water=4186
J/(kg·∘C),cwater=4186 J/(kg·∘C), and ?steam=2009
J/(kg·∘C),csteam=2009 J/(kg·∘C), respectively.
If the chamber initially contained 6.1906.190 moles of steam
(water) molecules before the ice was added, calculate the final
temperature ?fTf of the puddle...

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