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## Homework Statement

A 300-g aluminum vessel contains 200 g of water at 10

^{0}C.

100 g of steam at 100

^{0}C 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

^{0}C.

specific heat of water = 4190 J/kg

^{0}C.

specific heat of steam = 2108 J / kg

^{0}C.

## The Attempt at a Solution

I am using the [tex]\left|[/tex]Q

_{loss}| = [tex]\left|[/tex]Q

_{gained|}

[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) (100

^{0}C - T

_{eq}) = [(0.3kg) (910) (T

_{eq}- (10

^{0}C )) ]+ [(0.2kg) (4190) ((T

_{eq}- 10

^{0}C) ]

**after solving the equation, I have got T**

However, according to the solution, the answer should be T

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!

_{eq}= 24.28^{0}CHowever, according to the solution, the answer should be T

_{eq}= 100^{0}C. [\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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