Thermo Dynamics Help: Solving for Final Temperature and Total Mass of Ice Melted

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SUMMARY

The discussion centers on calculating the final temperature and total mass of ice melted when a 50g ice cube is introduced into a thermally insulated container with 200g of water at 25°C. The final temperature after the first ice cube is added is determined to be 2.5155°C using the equation Q1 + Q2 + Q3 = Q_water. For the second ice cube, participants suggest recalculating the heat exchanges based on the new initial temperature and considering the heat required to melt the ice without allowing the water temperature to drop below 0°C. The total mass of ice melted will depend on the heat lost by the water as it approaches 0°C.

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rdn98
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A single 50g ice cube is dropped into a thermally insulated container holding 200g of water. The water is initially at 25oC and the ice is initially at -15oC.

**********
a) What is the final temperature of the system after is has come to thermal equilibrium ?
b) Now let's drop a second 50gm cube of ice into the system. What is the final temperature of the system after it has come to thermal equilibrium for this second time?
c) What is the total mass of ice melted during this entire process?

part a)
Basically I figured out total heat absorbed by ice, which is equal to total heat lost by the water. With some algebra, I solved for Temperature, and the new temperature is now 2.5155 C.


Q1=m*c*delta T
= .05kg*specific heat ice*15 C
=1665J
Q2= L*m
= 333,000J/kg*.05kg =16650 J
Q3= mass*specific heat water*delta T
0.05kg*4190*T
= 209.5T
Q1+Q2+Q3=Q_water
where Q_water = .2*4190*(25-T)

T=2.5155 degrees C

part b) Okay, the final temp is now the initial temp for my water system..How do I change my equations to reflect the addition of the 2nd ice cube?

part c)
How do I go about finding total mass of ice melted?
 
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part b: Do exactly the same thing you did in part 1 using the new intitial temperature. Assuming that the second ice cube is also at 15 degrees, find the heat necesary to bring the ice up to 0 degrees and calculate how low that will bring the temperature of the water. From the wording of the question, I suspect if you calculate the heat necessary to melt all of the ice you will find that the water losing that much heat would drop below 0! That couldn't happen, of course, without freezing all of the water. Instead, calculate the heat the water would lose going to 0 and see what mass of ice that heat will melt. That will give an answer to (b) of 0 degrees and give you the answer to (c) as well.

Of course, if your teacher is being "cute" and you find that the ice will all melt without the water going to 0 degrees, then the final temp of the water is whatever you got that way, and the answer to (c) is "all 50 grams".
 
Don't forget to consider the new mass of the water for part b (different than 200 g).
 

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