# Thermal equilbrium question

1. Oct 5, 2014

### jonan_evans

Two 50g ice cubes are dropped into 200g of water in a glass. The water was initially at a temperature of 25°C, and the ice came directly from the freezer at -15°C. Neglect the heat capacity of the glass and any heat transfer to the environment. What is the final temperature of the drink when it arrives at thermal equilibrium? Is it all ice, all liquid water,or a mixture of liquid water and ice?

Please write a few lines describing the approach and identifying the concepts you will use to solve the problem.

Here are some useful constants:
R = 8.31JK−1mol−1
c of water = 4.2Jg−1K−1
c of ice = 2.1Jg−1K−1
density of water = 1g/mL
Lf for ice = 334J/g
K of ice = 2.2Wm−1K−1
σ = 5.67x10−8Wm−2K−4
CV,diatomic = 5/2 R
CV,monatomic = 3/2 R

2. Oct 5, 2014

### BvU

Hi Jonan, and welcome to PF. Something went wrong, because I miss items 2 and 3 from the template. You did a good job on 1, now indulge everyone who would like to help continuing with 2 and 3:
(you'll need them anyway as part of your answer: they ask for a description of approach and concepts!
But what's more: in PF using the template is mandatory. Read the guidelines to find out why it's so useful.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution

3. Oct 9, 2014

### rainleaf

Hi, I actually have the same problem so I'm going to try at items 2 and 3 ;)

2. Q = mcT and Q=mL

3. I reasoned that heat flows from the water to the ice cube, and the heat required to get the ice cubes to 0°C (used specific heat of water) and to melt them to all liquid is
(100g)(2.1J/gK)(15 K) + (100g)(334 J/g) = 3655 J

The water would've lost this heat. Use specific heat of water
3655 J = (200g)(4.2J/gK)(deltaT)
change in T = 4.35 K
25°C - 4.35°C = 20.65°C

So I would have 200g of water at 20.65 °C and 100g of water at 0°C. Of course heat would flow from hot to cold, so I found the average T:
100g of water at 0°C
200g of water at 20.65°C
300g of water with a third at 0°C and two thirds 20.65°C

(1/3)(0) + (2/3)(20.65) = 13.8°C; it is all water

This is what I have but I'm a bit unsure about the last step. I know to reach thermal equilibrium there must be no more heat flow, but could there be a way to compare rates of heat flow and make that equal to 0? Thanks!

Last edited: Oct 9, 2014
4. Oct 9, 2014

### nasu

100*334=33,400

5. Oct 9, 2014

### rainleaf

Using that correct calculation, to both change the ice to 0°C and melt it all would be:
(100(2.1)(15) + (100)(334) = 36550 J
This would correspond to a heat loss in the water:
36550 = (200)(4.2)(ΔT)
change in T = 43.5 K
This is too large a ΔT; the water would drop below freezing and even be below the T of the ice cubes.
This may mean the thermal eq is reached at a T below freezing.

If we say the water loses heat and goes to 0°C:
Q = (200)(4.2)(25) = 21000 J
This would correspond to a heat gain in the ice:
Q = (200)(2.1)(ΔT)
ΔT = 100K
This is too great a T change in the ice; it would even be above the T of the water

.. So thermal equilibrium is reached at 0°C, and it is a mixture of ice water.

-----> I'm not sure if this logic above makes sense in ruling out all ice melted or all water frozen.

But continuing...

If we have 100g of 0°C ice and 200g of 0°C water, the water will still transfer heat energy to the ice - as a liquid, it has more energy. This goes towards melting the ice (the ice will remain at 0°C assuming uniform heat transfer) and/or, freezing the water. I think this happens simultaneously because heat flows both ways.

I want to use Q = mL but if I think of heat gained by ice = heat lost by water, then I'm looking for 2 masses.
From that, I get

mcΔT + ΔmwaterL = mcΔT + ΔmiceL
(200)(4.2)(25) + Δmwater(334) = 100(4.2)(15) + Δmice(334)

from which I can get
Δmwater - Δmice = 53.44 g

But I'm not quite sure of this intuitively so:
I think that if 53.4g of ice gets melted and 0g of water gets frozen, that would be the difference in masses melted between the ice and water. (net gain 53.4 g water)

I think if 100g of ice is melted, and 46.6g of water is frozen, then we get a net gain 100-46.6 = 53.4g of water.

I think that means there is a net gain of 53.4g of ice turned into water.

100-53.4 = 46.6g ice left
200+53.4 = 253.4 water

The mixture is ice and water at 0°C, 46.6g of ice and 253.4g of water.

Last edited: Oct 9, 2014
6. Oct 9, 2014

### Staff: Mentor

I find that a reasonable approach for these sorts of problems is to do it in stages. First determine how much heat energy it would take to raise the ice to the freezing point (so ice at 0C). Then determine how much energy the liquid water has available before some of it would start to freeze. In other words, how much energy can come from bringing the liquid water from its initial temperature to 0C.

If the available energy exceeds that required to warm the ice to the freezing point then do so. This will leave you with ice at 0C and water at some new temperature. This is a new "starting point".

At this point ice can only melt via the heat of fusion. So once again determine how much heat energy it would take to melt all of the ice, and how much energy the water has left to give up before its all at 0C (same as the ice temp). Melt whatever amount of ice the available heat from the water allows.

Last edited: Oct 9, 2014
7. Oct 9, 2014

### rainleaf

Thank you so much! I really like how that approach allowed me to check what process was happening based on available energy. I was having difficulty thinking of loss of energy from the water as 'available energy' for the ice because I always thought of mc ΔT as heat loss, but for the ice, this is a energy gain/energy available towards bringing to a T or melting. Breaking it apart into all the steps really helped me to check what was happening. Thanks! I really learned a lot from this problem. :)