# Supercooled water

gianeshwar
My argument is that heat lost by lump of mass converted to ice will be gained by remaining water. But how do I find heat gained ?

Delta2

Are you familiar with the concept of heat of fusion? The freezing ice gives up its heat of fusion, which will warm the remaining water. The heat of fusion of water is given in the link.

Lnewqban
Mentor
The system is assumed to be insulated so that the change in internal energy from the initial state to the final state must be zero. Plus, the problem statement strongly implies that there will be both ice and liquid water present in the final state. What temperature do you think that mixture would be at?

gianeshwar
gianeshwar
The system is assumed to be insulated so that the change in internal energy from the initial state to the final state must be zero. Plus, the problem statement strongly implies that there will be both ice and liquid water present in the final state. What temperature do you think that mixture would be at?
This temperature cannot be found here. If I assume this temperature 0 degree,I get answer as 1/9 .But,I am doubtful.

How did you arrive at 1/9? Please show us your work. I think that you did not take into account the heat of fusion of water.

gianeshwar
How did you arrive at 1/9? Please show us your work. I think that you did not take into account the heat of fusion of water.
Let M is original supercooled water and m is mass of lump formed.
Heat lost 80 m
Heat gained(M-m)(0-(-10))
So m/M=1/9

Mentor
Let M is original supercooled water and m is mass of lump formed.
Heat lost 80 m
Heat gained(M-m)(0-(-10))
So m/M=1/9
The part that became ice also increased in temperature from -10 to zero.

jbriggs444 and gianeshwar
Mentor
The initial and final states of the system are:

Initial: M grams of liquid water at -10 C
Final: m grams of ice and (M-m) grams of liquid water at 0 C

If we take as the reference state for zero internal energy, liquid water at 0 C, then the internal energies of the initial and final states are:

Initial: ##U=M(-10-0)=-10 M\ calories##
Final: ##U=-80 m+(M-m)(0-0)=-80m\ calories##

So, $$-80m=-10M$$

This, of course, can also be done by taking as the reference state for zero internal energy, liquid water at -10 C. In this case,

Initial: ##U=M(-10-(-10))=0\ calories##
Final: ##U=m(0-(-10))-80m+(M-m)(0-(-10))=-80m+10M\ calories##

So,$$-80m+10M=0$$