Calorimetry Problem: Solving for Final Equilibrium Temp.

[a077456]

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 $$\left|$$Q_loss| = $$\left|$$Q_gained|
$$\left|$$( mc$$\Delta$$T ) _steam = $$\left|$$( mc$$\Delta$$T ) _Al + $$\left|$$( mc$$\Delta$$T ) _water |

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

after solving the equation, I have got T_eq = 24.28 ⁰C
However, according to the solution, the answer should be T_eq = 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!

 

[pat666]

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?

 

[a077456]

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 !

 

[a077456]

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.