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Here is the question from my textbook:

"In an insulated vessel, 250g of ice at $0^{\circ}$C is added to 600g of water at $18^{\circ}$C. What is the final temperature of the system?"

The solution says that it takes more heat to melt ice than to cool water, so some ice will remain. It says the final temperature should be $0^{\circ}$C, but no equation is given to explain that conclusion. Can the final temperature of this system be determined by reasoning alone?

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You need to consider the heat capcity of liquid water and the latent heat of fusion of water, and calculate whether the amount of heat removal needed to cool the water to 0 degrees is more or less than the amount of heat needed to melt the ice.

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Latent heat of melting of ice is approximately 80 calories. Heat "available" in 600g of water at 18C is 600x18 calories. Since $$250\cdot 80 > 600\cdot 18$$

Not all the ice can melt (unless external heat is added). Conclusion - the system reaches equilibrium at 0C at which point $$250-\frac{600\cdot18}{80}=115$$ grams of ice remain

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  • $\begingroup$ As a side note: if you did this in a sealed container you may find it drops to 0.1 degrees (Exactly) as you would have created a triple point cell en.wikipedia.org/wiki/Triple_point $\endgroup$
    – DarcyThomas
    Commented May 7, 2014 at 4:21
  • $\begingroup$ @DarcyThomas - you mean $0.01^\circ C$ - that is the triple point of water (by definition), not $0.1^\circ C$ $\endgroup$
    – Floris
    Commented May 7, 2014 at 13:44
  • $\begingroup$ @Floris Oops yes 0.01∘C My bad. $\endgroup$
    – DarcyThomas
    Commented May 7, 2014 at 19:15
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One way to answer this type of question, where you don't how much (some or all) of the ice melts, is to bring all the various components to the same arbitrary final state, and then see what has to happen to maintain zero heat flow.

Lets assume that the final state is all liquid water at $0^o$ C. (Actually, the least likely result, but a convenient one!)

As others have explained, to melt all the ice, you need to add $$250\times 80=20,000 \text{ calories}$$

To cool the warm liquid water to $0^o$ C, you need to remove:$$600\times 18= 10,800\text{ calories}$$So, now you have $850$ grams of ice cold water. The only remaining problem is to restore the condition of zero heat energy flow.

If you needed to add energy, you could find out how much that energy would warm the water. In this case, you can find out how much liquid water would freeze when you removed $9200$ calories.

The method can be extended to sub-zero ice and even some super-heated steam; bring all the different states of water to one mass of water, then move it as a single "lump" wherever you need to restore zero heat flow...

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  • $\begingroup$ So is this correct: First we compare the amount of heat needed to melt all of the ice with the amount of heat that must be removed from the water to cool it to $0^{\circ}$C. We see that not all of the ice needs to melt to cool the water to that temperature. Therefore, some ice will remain when the system reaches equilibrium. And the equation we are using is $Q_{ice}=-Q_{water}$, where $Q_{ice}=L_{fusion}m_{initial}$. $\endgroup$
    – curiousGeorge119
    Commented May 7, 2014 at 13:53
  • $\begingroup$ Well, I wonder if it is more correct to say that $Q$ is the heat absorbed by the ice, which would be exactly equal to the heat released by the water? $\endgroup$
    – curiousGeorge119
    Commented May 7, 2014 at 14:03

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