## 4C.13

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AlyssaYeh_1B
Posts: 100
Joined: Sat Aug 17, 2019 12:16 am

### 4C.13

An ice cube of mass 50.0g at 0.0ºC is added to a glass containing 400.0g of water at 45.0ºC. What is the final temperature of the system? Assume that no heat is lost to the surroundings.

How would this problem be solved? I understand that we need to use the enthalpy of fusion and the specific heat of water, but I'm unsure of how to start.

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
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### Re: 4C.13

The heat gained by the water in the ice cube will be equal to the heat lost by the initial sample of hot water. The enthalpy change for the water in the ice cube will be composed of two terms: the heat to melt the ice at 0 celsius to the final temperature.
heat (ice cube) = (50g/18.02 g.mol^-1)(6.01 x 10^3 J.mol^-1) + (50.0g)(4.184 J.celsius^-1.g^-1)(Tfinal - 0 celsius)
heat (water) = (400g)(4.184 J.celsius^-1.g^-1)(Tfinal - 45 celsius)
Set these two equal to each other, then solve for Tfinal.

MaggieHan1L
Posts: 100
Joined: Thu Jul 11, 2019 12:17 am

### Re: 4C.13

Whenever there is phase change you add the heat of fusion/vaporization to the amount of energy it takes to raise the substance a certain amount of degrees. If you look at the phase change curve you can see it's flat then it starts increasing in temperature. The flat area is the phase change.

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