Final temperature after pouring soda into glass with ice...

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  • #1
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Homework Statement


Soda from a ms=12 oz can at temperature Ts=13 degrees C is poured in its entirety into a glass containing a mass mI=0.16kg amount of ice at temperature TI=-19.5 degrees C. Assume that ice and water have the following specfic heats c1=2090 J/(kg*C) and cs= 4186 J/(kg*C), and latent heat of fusion of ice is Lf=334 kj/kg.

a) In degrees celsius, what is the final temperature Tfinal of the mixture?
b) Write an expression for how much of the ice mmelted has melted?
c) In kilograms, how much of the ice mmelted has melted?

Homework Equations


Q= m*c*deltaT

The Attempt at a Solution


So I think you start by converting the 12 oz can to kg to get the mass at that particular temperature. Then would you set two separate Q equations equal to one another to solve for Tf? So Q= ms*cs*(Tf-13) and Q= mI*cI*(Tf-(-19.5)) and set these equal to each other, but then how does the latent heat of fusion of ice play into the calculation?
 

Answers and Replies

  • #2
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Ok

Homework Statement


Soda from a ms=12 oz can at temperature Ts=13 degrees C is poured in its entirety into a glass containing a mass mI=0.16kg amount of ice at temperature TI=-19.5 degrees C. Assume that ice and water have the following specfic heats c1=2090 J/(kg*C) and cs= 4186 J/(kg*C), and latent heat of fusion of ice is Lf=334 kj/kg.

a) In degrees celsius, what is the final temperature Tfinal of the mixture?
b) Write an expression for how much of the ice mmelted has melted?
c) In kilograms, how much of the ice mmelted has melted?

Homework Equations


Q= m*c*deltaT

The Attempt at a Solution


So I think you start by converting the 12 oz can to kg to get the mass at that particular temperature. Then would you set two separate Q equations equal to one another to solve for Tf? So Q= ms*cs*(Tf-13) and Q= mI*cI*(Tf-(-19.5)) and set these equal to each other, but then how does the latent heat of fusion of ice play into the calculation?
O

Homework Statement


Soda from a ms=12 oz can at temperature Ts=13 degrees C is poured in its entirety into a glass containing a mass mI=0.16kg amount of ice at temperature TI=-19.5 degrees C. Assume that ice and water have the following specfic heats c1=2090 J/(kg*C) and cs= 4186 J/(kg*C), and latent heat of fusion of ice is Lf=334 kj/kg.

a) In degrees celsius, what is the final temperature Tfinal of the mixture?
b) Write an expression for how much of the ice mmelted has melted?
c) In kilograms, how much of the ice mmelted has melted?

Homework Equations


Q= m*c*deltaT

The Attempt at a Solution


So I think you start by converting the 12 oz can to kg to get the mass at that particular temperature. Then would you set two separate Q equations equal to one another to solve for Tf? So Q= ms*cs*(Tf-13) and Q= mI*cI*(Tf-(-19.5)) and set these equal to each other, but then how does the latent heat of fusion of ice play into the calculation?
Okay so I did: 4186*0.340194*(13-T) = 2090*0.16*(T+19.5)
and found Tf=6.82 and the system says this is incorrect but the hint says "You may have forgotten that ice experiences phase change at 0 degrees celsius. Take into account the latent heat.
As stated before, I'm not sure how to incorporate this into my solution.
 
  • #3
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I haven't gone through the numbers, but because of the wording of the question, I suspect this is what is happening in this problem.
The liquid gets added to the ice. The liquid warms up all of the ice to its melting point. Then the liquid melts all of the ice into a liquid. Then those two liquids reach some equilibrium temperature. Now it could be the case that the liquid does not have enough energy to do all of that. But that should come out in the numbers. So yes, you have to remember that the ice changes phase at 0C and requires energy from the warm(er) liquid to do that. And the amount of energy required to warm up 1 kg of ice 1 degree C is not the same amount of energy required to warm up 1 kg of water 1 degree C, which is not the same energy required to melt 1 kg of ice.

Edit: I defer to @Chestermiller. He is right (and I was wrong) that the problem statement seems to imply that not all of the ice is melted. I guess it would help if I read the whole problem carefully. :(
 
Last edited:
  • #4
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The problem statement (and your calculation) seems to imply that not all the ice will have melted when the system has reached thermal equilibrium. If that is the case, what is the final temperature of the system?
 
  • #5
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Any more input about this problem? I still can't seem to get the right answer
 
  • #6
haruspex
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Any more input about this problem? I still can't seem to get the right answer
Things will happen in stages, and one simple, if laborious, approach is to model those.
1. Ice warms and liquid cools until one of them reaches 0C.
2. If the water reaches 0C first, it will start to freeze, giving out latent heat.
2.1 If all the water freezes and the original ice is still less than 0C, the two will balance out at some temperature below freezing.
2.2 If the ice reaches 0C before all the water freezes you are done.
3. If the ice reaches 0C first then it will start to melt,
etc.
 

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