Calorimetry Problem: Solving for Final Equilibrium Temp.

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Homework Help Overview

The problem involves a calorimetry scenario where an aluminum vessel contains water, and steam is introduced into the system. The objective is to determine the final equilibrium temperature of the system after the steam interacts with the water and aluminum.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the energy balance between the steam, water, and aluminum, questioning the setup and whether to account for phase changes. There is consideration of latent heat and how it affects the final temperature.

Discussion Status

Some participants have offered insights regarding the latent heat of steam and its conversion to water. There is an ongoing exploration of the implications of phase changes on the final temperature, with multiple interpretations being considered.

Contextual Notes

Participants note the importance of accounting for latent heat in the calculations and express uncertainty about the state of the steam after the interaction. There is a lack of consensus on whether all steam condenses or if some remains in the gaseous state.

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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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