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So here is my doubt -

10g of ice at -10C is added to 10g of water at 85C. What is the final temperature and amount of ice left in the system? (System is kept inside an ideal insulator)

This question may seem quite easy for many of the readers, but please help me with such questions as I find them tricky. I understand that this is a question which involves the application of the Q=mc(delta T) formula and that we have to equate the Q's of both ice and water in order to obtain the answer.

Another thing which is troubling me is that the answers to most of such questions that I have encountered is 0C, and this question is not an exception.

Please suggest me on how to go about such questions, and provide some conceptual clarity on the topic if possible. Thanks.

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  • $\begingroup$ Solve this in steps rather than looking for one equation to cut the Gordian knot. So how much heat does it take to raise the ice to 0 C, and when you extract that much heat from the water what is the temperature of the water? Then solve the next part of the problem. $\endgroup$ – MaxW Feb 10 '17 at 18:22
  • $\begingroup$ I think the reason that the answer to these kind of questions are so often 0$^o$C is to 1) drive home the point of how much latent heat is involved and 2) to be sure that you see that any ice remaining means 0$^o$C. $\endgroup$ – airhuff Feb 11 '17 at 0:30
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In this type of question u should bring all the matter into one state and calculate net heat

In this solution we will consider heat released as + ve heat absorbed as - ve and the state to which we will bring all the matter be water at 0 ° c

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Here we were left with 20g of water at 0°c and zero heat if heat was some positive quantity we would use the heat to raise 20g water to a specific temperature

By equation 20×T×1=heat left T is temp to which 20g water is raised

if heat was negative quantity we would convert part of water at 0°c to ice at 0°c by equation

M×80=heat left M mass of water which is converted to ice

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  • $\begingroup$ I'll point out that textbooks now tend to use Joules not calories. $\endgroup$ – MaxW Feb 10 '17 at 18:58
  • $\begingroup$ It will just different in value of latent heat and specific heat process will be exactly same $\endgroup$ – user41111 Feb 10 '17 at 19:03
  • $\begingroup$ I don't disagree with your method or answer. I'm just trying to point out that "old" books use calories for such problems and "new" books use Joules. I'm an old dog and would use calories too... $\endgroup$ – MaxW Feb 10 '17 at 19:05
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    $\begingroup$ OK in place of 1 cal we can place 4.2J as 1 cal= 4.2J $\endgroup$ – user41111 Feb 10 '17 at 19:07

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