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# Specific Heat Capacity watch

1. Hi,

So currently working through some questions and wanted some clarification over why what I did was wrong.

The question is: calculate the temperature is 10cm^3 of 0.1mol dm^-3 NaOH is reacted with 20cm^3 of 0.1mol dm^-3 HCl. We're told that the enthalpy change of neutralisation is -58kJmol^-1.

So what I did is:
n=cv
n=0.1x30
n=3mol
3/1000=0.003
0.003x-58000=-174
then rearranged mcΔt to
Δt=-174/4.18x30 to get -1.39C however that isn't correct. Can anyone tell me why and where I went wrong? The answer in the textbook is 0.46°C.

Thanks.

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