Calorimetry Problem: Solving for Final Equilibrium Temp.

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The discussion revolves around calculating the final equilibrium temperature of a system consisting of an aluminum vessel, water, and steam. The initial setup involves a 300-g aluminum vessel with 200 g of water at 10°C and 100 g of steam at 100°C. The user initially calculated the equilibrium temperature as 24.28°C but later realized the importance of considering the latent heat of steam as it condenses into water. The consensus is that during the phase change, the temperature remains at 100°C until all steam has converted to water. The user ultimately understands the necessity of accounting for both the latent heat and the mass of steam condensed in their calculations.
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Homework Statement


A 300-g aluminum vessel contains 200 g of water at 10 0C.
100 g of steam at 1000C is poured into the container, what is the final equilibrium temperature of the system?


Homework Equations


Given that the:
specific heat of Al = 910 J/kg 0C.
specific heat of water = 4190 J/kg 0C.
specific heat of steam = 2108 J / kg0C.


The Attempt at a Solution



I am using the \left|Qloss| = \left|Qgained|
\left|( mc\DeltaT ) steam = \left|( mc\DeltaT ) Al + \left|( mc\DeltaT ) water |

(0.1kg) ( 2108) (1000C - Teq) = [(0.3kg) (910) (Teq - (100C )) ]+ [(0.2kg) (4190) ((Teq - 100C) ]

after solving the equation, I have got Teq = 24.28 0C
However, according to the solution, the answer should be Teq = 100 0C. [\b]

I am just wondering how should the setup be? Do I have to consider that the steam is actually turning into ice or ice turning into steam? Therefore, should we also use the Q = mL where L is the latent heat; m = mass and Q = heat?

THank you very much and I am looking forward to hear any reply!
 
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Your'e forgetting about the latent heat of fusion from steam to water but still I wouldn't have thought it would equalize at 100 degrees. And what ice are you talking about?
 
I think I made a mistake in that part. I was trying to say that the steam turn into water. So which would be the case: 1) all water 2) Water + Steam left over in the final stage?

I have tried to encounter with the Latent heat part, and it will be as follow:

Let L of steam = 2108

| (mc (delta T) + mL) of steam | = | mc (delta T) of Al + mc (delta T) of water |
(0.1kg) ( 2108) (100 C - 100 C) + m (2108) = [(0.3kg) (910) (Teq - (100C )) ]+ [(0.2kg) (4190) ((Teq - 100C) ]

Then at this point, I would have 2 unknowns, which is the mass of steam condensed, and the Teq, which is what I am looking for...

How should I approach this problem? Thank you very much !
 
I have understand how to do this problem now. Thanks!
This is because at the phase changes stage, steam is converting into water, therefore the temperature must stay at 100 degree.
 
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