Problem statement: A mixture containing 500[g] water and 100[g] ice is initially in thermal equilibrium at a temperature of 0[°C] in an insulated vessel. Then 200[g] of water vapor at 100[°C] is introduced into the vessel. Assume that the heat capacity of the vessel is negligible, further consider: Lmelting = 3.33*105[J/kg], Lvaporization = 22.60*105[J/kg] and specific heat of water cwater = 4200[J/kg*°C]. Determine the final composition of the system in grams and the final temperature.

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Chapter1: Units, Trigonometry. And Vectors
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Problem statement: A mixture containing 500[g] water and 100[g] ice is initially in thermal equilibrium at a temperature of 0[°C] in an insulated vessel. Then 200[g] of water vapor at 100[°C] is introduced into the vessel. Assume that the heat capacity of the vessel is negligible, further consider:
Lmelting = 3.33*105[J/kg], Lvaporization = 22.60*105[J/kg] and specific heat of water cwater = 4200[J/kg*°C].
Determine the final composition of the system in grams and the final temperature.

I do not understand the relationship between the percentage of condensed water vapor and the percentage of total water (0.6 kilograms) at 100 degrees that evaporates. [1-x] and [x] respectively.

 

Balance the Heats:
[Head
Heat
gain!
(Heat
[Had gon by] + [tid gain by] + [ord gouin by] - [thed Jout be out by
= ["
Head lost by kg
day vapor a 100°C
ice
kg water
10.6 kg wet
Qice + Quaker + mix xxL
vaporise
0,333x 105 +2.52x10² + 0.6××× 2².6×10²
i. x = 0.0922 ≈ 0.1
mx (1-x] xLvaporise
= 0.2 × [1-x] x 2².6 X 10³
=
Transcribed Image Text:Balance the Heats: [Head Heat gain! (Heat [Had gon by] + [tid gain by] + [ord gouin by] - [thed Jout be out by = [" Head lost by kg day vapor a 100°C ice kg water 10.6 kg wet Qice + Quaker + mix xxL vaporise 0,333x 105 +2.52x10² + 0.6××× 2².6×10² i. x = 0.0922 ≈ 0.1 mx (1-x] xLvaporise = 0.2 × [1-x] x 2².6 X 10³ =
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