Ice/water thermodyamics confusion

[luke850]

Homework Statement

If you mixed 0.2kg of ice that is at -5 degrees C with 0.02kg of water that is at 15 degrees C, what will be the temperature and condition of the final state once equilibrium is achieved?

Homework Equations

Q=mL
Q=mc(delta T)

The Attempt at a Solution

After setting up a heat lost, heat gained equation, I can see how many joules are lost/gained by each component. However, I am unable to apply this information to find the final temp.

Q ice= (0.2kg)(333000J/kg)+(0.2kg)(2090J/kg*K)(5K)= 68690J
Qwater= (0.02kg)(333000J/kg)+(0.02kg)(4190J/kg*K)(-15K)=-7917J

Then I thought...

delta T for ice= [-7917J - (0.2kg)(333000J/kg)]/(0.2kg)(2090J/kg*K)

but this answer doesn't seem plausible.

 

[danago]

You should be thinking about whether the final equilibrium system will be all liquid water, all ice or a mixture of ice and liquid water.

Calculate the amount of energy loss required to bring the water from 15 degrees down to 0.

Calculate the amount of energy gain required to bring the ice from -5 degrees up to 0 degrees.

What does this tell you?

 

[Delphi51]

Welcome to PF, Luke.
You started with heat lost by water = heat gained by ice, right?
On the left side you should have a (15-T) where T is the final temperature.
On the right you should have a (T- -5). So you can't find a number for either the heat lost or the heat gained with the unknown T in there. However, you can solve the equation to find T.

I suppose on the first try you will make the simplest assumption - that the ice does not melt. If working that out gives a T greater than zero, you guessed wrong and need to assume that the ice melts - meaning you need and mcΔT term plus a heat of fusion term. Yikes, this could get complicated! I hope you will show your calcs so we can see how it works out.

Edit: oops, simultaneous post. Better follow Danago's idea, it is a lot clearer!

 

[luke850]

Thanks for the prompt reply. I elaborated further on the work I've done- I think it tells me that there isn't nearly enough heat going into the ice to melt all of it. However, I'm not sure where to go from here.

 

[luke850]

Ok, so I tried again, this time omitting the heat of fusion terms, and the answer seems reasonable. Still not sure if it's correct, and more importantly, why heat of fusion terms aren't necessary.

(0.2kg)(2090J/kg)(T-(-5))=(0.02kg)(4190J/kg*K)(15-T)
T= -0.401 degree Cand going back further...
Ice to 0 degrees C=(0.2kg)(2090J/kg*K)(5K)=2090J
water to 0 degrees C= (0.02kg)(4190J/kg*K)(15K)=1257J

so i am now sure that the ice does not melt! still unsure about the accuracy of my final temp calculation though...

it is clear to me now why a heat of fusion term wouldn't be necessary for the ice side of the equation, but how about the water side?

 

[danago]

If all of the ice is to increase in temperature to its melting point, it will require:
<br /> Q_{ice} = (0.2)(2090)(5) = 2090\;J<br />

If the water is to cool down to its freezing temperature, it will release:
<br /> Q_{water} = (0.02)(4190)(15) = 1257\;J<br />

Based on these figures, you can conclude that the sensible heat (i.e. associated with a temperature change) released by the water is not enough to bring the ice to its melting point. So basically, the water will reach its freezing point but the ice will be somewhere in between -5 and 0 degrees. What will this temperature be?

Since the ice will still be at a temperature that is lower than the water at 0 degrees, heat will still spontaneously flow from the water to the ice. This will cause the ice to increase in temperature while the water freezes (thus releasing latent heat).

 

[luke850]

I posted the temp I think it might be on my previous post, but I have no way of knowing whether it's accurate, or whether the equation is remotely correct. Furthermore, it seems that every time I try to calculate a temp using latent heat on either (or both) sides of the equation, the change in temp quickly becomes very unreasonable.

For example:

(.2)(2090)T=(0.02)(4190)(15)+(0.02)(333000)
T=21.95

 

[danago]

I posted the temp I think it might be on my previous post, but I have no way of knowing whether it's accurate, or whether the equation is remotely correct. Furthermore, it seems that every time I try to calculate a temp using latent heat on either (or both) sides of the equation, the change in temp quickly becomes very unreasonable.

For example:

(.2)(2090)T=(0.02)(4190)(15)+(0.02)(333000)
T=21.95

As we have already calculated, as the water cools down to 0 degrees, it loses 1257J of energy. This corresponds to an increase in temperature of the ice:

<br /> \Delta {\rm T} = \frac{Q}{{mc}} = \frac{{1257}}{{(0.2)(2090)}} = 3<br />

Therefore, the ice will be at around -2 degrees once the water has reached its freezing point.

The water will then begin freezing as it releases more heat, thus increasing the temperature of the ice. You will next need to consider whether the water will be able to completely freeze before the ice starts melting.