Calorimetry Problem: Solving for Final Equilibrium Temp.

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The discussion centers on calculating the final equilibrium temperature of a system containing a 300-g aluminum vessel, 200 g of water at 10°C, and 100 g of steam at 100°C. The specific heats used are 910 J/kg°C for aluminum, 4190 J/kg°C for water, and 2108 J/kg°C for steam. The initial calculations suggested a final temperature of 24.28°C, but the correct equilibrium temperature is 100°C due to the latent heat of steam condensing into water. The key takeaway is that during phase changes, the temperature remains constant, necessitating the inclusion of latent heat in the 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!
 
Last edited:
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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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