[Blueberries116]

The problem is as follows:

A flask whose heat capacity is negligible is filled with [itex]20[/itex] grams of ice at [itex]-80^{\circ}C[/itex] and [itex]7[/itex] grams of water at [itex]80^{\circ}C[/itex]. Find the composition of the mixture in the flask when the equilibrium is reached.

The alternatives given are as follows:

[itex]\begin{array}{ll}
1.&\textrm{2 g. of water and 25 g. of ice}\\
2.&\textrm{5 g. of water and 22 g. of ice}\\
3.&\textrm{7 g. of water and 20 g. of ice}\\
4.&\textrm{3 g. of water and 21 g. of ice}\\
5.&\textrm{4 g. of water and 23 g. of ice}\\
\end{array}[/itex]

For this problem I'm not sure how to proceed. What I've attempted to do was to assume that the warmer water will lose heat and give it to the ice so it will melt. But I don't know exactly how to translate this into an equation.

So what I did was to add the heat to warm up the ice and melt the ice with that of the cooling of the water.

[itex]mL_{f}=q_{water}[/itex]

[itex]80m=7\left(1\frac{cal}{g^{\circ}C}\right)\times 80[/itex]

[itex]m=7[/itex]

But I don't know exactly if this is the right way to approach this problem. Can someone help me with what would be the right concept?

[AWK]

You need to make a balance of heat absorbed and released.
Ice absorbs heat to warm up to 0°C and partially melts, and water releases heat to cool to 0°C. The main problem is the accuracy of the calculations. While the specific heat of water at these temperatures changes slightly, the specific heat of ice changes by about 30%. The authors of this problem should specify possible simplifying conditions.

[Blueberries116]

You need to make a balance of heat absorbed and released.
Ice absorbs heat to warm up to 0°C and partially melts, and water releases heat to cool to 0°C. The main problem is the accuracy of the calculations. While the specific heat of water at these temperatures changes slightly, the specific heat of ice changes by about 30%. The authors of this problem should specify possible simplifying conditions.

I'm still unable to do the heat balance on my own. How does the equation in the end would be like?.

Would it be like this?

[itex]mL{f}+mc\Delta T = mL{f}[/itex]

or would it be like

[itex]mL{f}+mc\Delta T = -mc \Delta T[/itex]

Can you guide me exactly?. I don't know exactly what quantities should be put in the equation.

I do know that the latent heat of fusion for ice is 80 cal and the specific heat for ice is [itex]0.5 \frac{cal}{g\cdot ^{\circ}C}[/itex] and the specific heat for water is [itex]1\,\frac{cal}{g\cdot ^{\circ}C}[/itex]. Am I right with those?.

[AWK]

Only part of the ice will melt. It doesn't follow from the equation you wrote.
You need to calculate it, then subtract this mass from the ice mass and add to the mass of water to get the final result.

specific heat for ice is 0.5 cal/(g·°C)

This is what I meant by simplifying conditions.

[Borek]

Note: in this type of problems it is often not possible to write the equation describing whole process before checking details.

It is obvious at the beginning hot water will cool down and the ice will heat up. Depending on their relative amounts the end result can be either just liquid water with T≥0°C, just a solid with T≤0°C, or an ice/water mixture with T=0°C. Each of these cases is described by a slightly different equation and you can't say beforehand which one of the formulas is the right one*

The best approach is to do the problem stepwise. First - when you mix cold ice and hot water ice heats up, water cools down. Find out how much heat is required to heat up the ice to the melting point. Found out how much heat is the water capable of giving away. Is there enough heat to heat up ice to 0°C? If so, it was heated and now you have ice ready to melt and cooled down water (you don't need to find its temperature, it is enough to just subtract the heat used to heat up the ice from the amount of heat water had). If there was not enough heat to heat up the ice, you have some cold ice (T<0°C) and water ready to freeze. Now proceed further - depending on which case you are dealing with either ice will melt, or water will freeze, and you have to again compare energies involved to find out how will the process end.


*actually you can describe the ice/water mixture with the system of equations, including not only energy balance, but also mass balance, then the equations will be the same each time, but it doesn't make finding the solution much easier, as the math becomes more involved

[Blueberries116]

The only way how I made it to work was to think this:

the heat needed to warm up the ice to be zero celcius will be:

[itex]q=mc\Delta T = (20)(0.5)(80)=800\,cal[/itex]

The heat needed to cool down the hot water:

[itex]q=(7)(1)(80)=560\,cal[/itex]

Then it means that there is not enough heat to melt the ice.

Some of this remaining heat will be transformed into ice:

[itex]800-560=mL{f}[/itex]

[itex]m=\frac{240}{80}=3\,g[/itex]

Then this is the amount of ice which will be added to the initial amount of ice, therefore.

[itex]20+3=23\,g[/itex]

and the amount of liquid which is initially at [itex]80^{\circ}C[/itex] will be lost as it is transformed into ice:

[itex]7-3=4\,g[/itex]

Therefore the answer would be the fifth option and this checks with my answers sheet.

[Borek]

The only way how I made it to work was to think this:

the heat needed to warm up the ice to be zero celcius will be:

[itex]q=mc\Delta T = (20)(0.5)(80)=800\,cal[/itex]

The heat needed to cool down the hot water:

[itex]q=(7)(1)(80)=560\,cal[/itex]

OK

Then it means that there is not enough heat to melt the ice.

True, but what is more important here is that there is not enough heat to even heat the ice up to 0°C. After this step you end with ice colder than zero and water cooled down to zero.

Some of this remaining heat will be transformed into ice:

[itex]800-560=mL{f}[/itex]

[itex]m=\frac{240}{80}=3\,g[/itex]

Then this is the amount of ice which will be added to the initial amount of ice, therefore.

[itex]20+3=23\,g[/itex]

and the amount of liquid which is initially at [itex]80^{\circ}C[/itex] will be lost as it is transformed into ice:

[itex]7-3=4\,g[/itex]

Therefore the answer would be the fifth option and this checks with my answers sheet.

Looks OK to me.