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For this multiple question:

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The correct answer is B. However, I don't understand.

You are halving the volume so for the original solution the number of moles of HCl is 0.1 moles and in the new solution, the number of moles of HCl is 0.05 if we are using the same concentration.

Therefore as there are less number of moles, won't the temperature increase (ie. energy release) get halved?

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  • $\begingroup$ Question is about temperature not heat. $\endgroup$ – Mithoron Jan 18 '16 at 22:48
  • $\begingroup$ @Mithoron, but isn't temperature the measure of the average heat energy? $\endgroup$ – CCC Jan 18 '16 at 22:51
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    $\begingroup$ LOL - But the heat released raises the temperature of the solution. You have half the reactants, but only half the amount of solution to heat too. $\endgroup$ – MaxW Jan 18 '16 at 22:57
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    $\begingroup$ @CCC You are correct. The answer to the question as worded is B. The person that created the question either solved it wrong or omitted the doubling of concentrations. $\endgroup$ – A.K. Jan 19 '16 at 0:44
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The correct answer is B in both the case the final concentrations are not changing. Therefore same will be the heat of rise. But heat released will be less but volume of the final solution also got halved. So temp. raise will be same.

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