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10 g of ice at −10 °C is added to 10 g of water at 85 °C. 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 0 °C, 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.

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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
  • $\begingroup$ Just from my memory remembering some numbers, it would result as all water 0 Deg C. $\endgroup$ – Poutnik Sep 21 '20 at 11:28
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In this type of question, you should bring all the matter into one state and calculate net heat.

In this solution, we will consider heat released as positive heat, absorbed as negative, 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 20 g of water at 0 °C and zero heat if heat was some positive quantity we would use the heat to raise 20 g water to a specific temperature.

By equation $20\times T\times1=\text{heat left}$
$T$ is temperature to which 20 g water is raised

If heat was negative quantity, we would convert part of water at 0 °C to ice at 0 °C by equation

$M\times80=\text{heat left}$
$M$ 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
  • $\begingroup$ Using photos/screenshots of text (especially if handwritten ) instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers. Additionally, it can be challenge to decipher. Consider copy/pasting or rewriting of the essential parts and using of MathJax for eventual formatting of mathematical/chemical formulas or equations. $\endgroup$ – Nisarg Bhavsar Jun 18 at 9:19

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