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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 [tex]\left|[/tex]Qloss| = [tex]\left|[/tex]Qgained|
[tex]\left|[/tex]( mc[tex]\Delta[/tex]T ) steam = [tex]\left|[/tex]( mc[tex]\Delta[/tex]T ) Al + [tex]\left|[/tex]( mc[tex]\Delta[/tex]T ) 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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