When 50 g of ice at 0 ℃ added to 10 g of steam at 100 ℃, 60 g of water is formed. What is the temperature of water?

Answer: 40.2 ℃.

I'm not asking for the workout of this question. I just want to know how can I reach to this answer, preferably with explanation. Thanks.

Heat absorbed=heat released, correct?


Well, first you'd like to transform your steam to water and see how many Joules you get from that (you'll have to look up the specific heat of vaporization of water in a textbook). Then melt the ice with that heat (here you'll need the specific heat of fusion) and heat the resulting water with the leftovers, if any (here you'll need the specific heat capacity of water). And finally, just mix together two amounts of water at different temperatures.

Upd. OK, I mostly agree with what you are doing, except for the last operation. You check the mass of ice you can melt with the heat obtained from the condensation, and found it to be 67.66 g. Good, but we don't have that much. We only have 50 g. What does that mean? It means you can't spend all that heat on melting. When you're done with melting, there is some leftover amount of heat. Now we use it to heat the water we obtained by melting. Can you do that?

  • $\begingroup$ Ivan what do you mean by ' heating the resulting water with the leftovers'?? Which is which? And why do we need to heat the 2 of them? And you've mentioned to ' mix together two amounts of water at different temperatures'. Meaning?? BTW I've did the first 2 parts and I've posted it on my question. I'm not sure whether I did it correctly. Can you have a look at it? $\endgroup$ – Theresa Oct 5 '15 at 10:58
  • $\begingroup$ Wait a minute, I'll edit my answer. $\endgroup$ – Ivan Neretin Oct 5 '15 at 11:36

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