Final temperature of ice and water [closed]

My answer doesn't seem likely but I can't see where I've gone wrong:

Q: Ice of mass 0.125kg at -5 degrees is added to water of mass 0.25kg at 8 degrees. Calculate the final temperature of the water at thermal equilibrium. Assume c of ice=2100J/kg°C, c of water=4200J/kg°C and $L_f$ water=334000J/kg

My attempt: heat to turn ice to zero + heat to melt ice + heat to raise 'melted ice' to x degrees = heat released by water to x degrees.

$(0.125*5*2100)+(0.125*334000)+(0.125*4200*x)=(0.25*4200*8-x)\\1312.5+41750+525x = 1050(8-x)\\43062.5+525x=8400-1050x\\1575x=-34662.5\\x=-22.00793651\\$

I was expecting a temperature above zero.

closed as off-topic by Jon Custer, Kyle Kanos, stafusa, Cosmas Zachos, ConifoldMar 21 '18 at 2:50

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These kind of problems must be considered this way: we don't know what actually happens, so I must check all the cases and the solution will be the one that has no contradiction.

You don't know the final temperature. Hence you cannot tell if all ice melts or not. Consequently, you have to consider both cases, and the one that has no contradiction is the correct one.

If you made the math correctly (whiuch I have not checked), it's obviously a contradiction. It's not only you "expected" $T_{final}>0$, you actually supposed it initially because you've used all formulas that are only applicable to that case.

Hence, you got a contradiction. The correct case must be the other one: not all ice melts down.

Let's call $m$ the mass of ice that actually melts down.

You again apply $\Delta Q=0$, which means, the sum of all heats (sign included) must be zero; in other words "heat gained = heat lost".

Heat for ice from -5ºC to 0ºC (btw, include FULL units). + Heat to melt down $m$ kilograms of ice = Heat lost by the water to pass from 8ºC to 0ºC.

Now your unknown is the mass of ice melt. The final temperature is not an unknown anymore, because if there's still ice, the temperature is for sure 0ºC.

• I think I did it (can't get it to format nicely): Energy released by water from 8ºC to 0ºC: $(4200J/kgºC*8ºC*0.25kg)=8400J$ Energy for ice to reach 0ºC: $(2100J/kgºC*5ºC*0.125kg)=1312.5J$ Energy to melt ice: $(334000J/kg*0.125kg)=41750J$ Total energy to heat and melt ice: $(41750J+1312.5J)=43062.5J$ $43062.5J>8400J$: Therefore, not enough energy released by water to melt all the ice, so water must be at 0ºC with m quantity of ice. Mass of ice that energy could melt: $1312.5J+(mkg*334000J/kg)=8400J\\ 334000mJ=7087.5J\\ m=0.02122005988kg\\ m≈21.2g$ – Paddington Mar 18 '18 at 22:50
• As other comments say, we are not here to check maths. I told you how this kind of problems are overcome. Now it's you who must check the math. It looks plausible, but check it carefully. – FGSUZ Mar 19 '18 at 11:50
• To the downvoters: commenting why is always helpful. I didn't give a solution, but instructions on how to do it, and they are correct. What's wrong? – FGSUZ Mar 19 '18 at 11:51