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A question in the book says :

A $25.95 g$ sample of methanol at $35.6 °C$ is added to a $38.65 g$ sample of ethanol at $24.7 °C$ in a constant pressure calorimeter If the final temperature pf the combined liquids is $28.5 °C$ and heat capacity of the calorimeter is $19.3 J/°C$ determine the specific heat of methanol? The heat capacity of ethanol is $2.46 J/(g\cdot°C)$

My solution was :

$q_{methanol}=-q_{ethanol}$

msT=-msT

$25.95g \cdot (28.5°C - 35.6°C) \cdot s = - 38.65g \cdot (28.5°C-24.7°C) \cdot 2.46 J/(g\cdot °C)$

$\therefore s = 1.96 J/(g\cdot °C)$

but the correct answer is $2.36 J/(g\cdot°C)$

and I think that there is a mistake because the heat capacity is given and I didn't use it ??

Could anyone explain what's wrong?

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  • $\begingroup$ It's a good idea to include units in your equation. Where does the number 38.5 come from? (I think you mean 28.5). Also, why are you multiplying the right side by 2.64? $\endgroup$ – Klik Jul 3 '14 at 12:20
  • $\begingroup$ yes i meant that.. 2.46 is the specific heat of ethanol I got it from the book $\endgroup$ – Maher Jul 3 '14 at 12:24
  • $\begingroup$ You don't mention ethanol in your problem, you only mention methanol. Could you double check the question you've written. Is it that you are given a sample of methanol and it is added to a sample of ethanol? $\endgroup$ – Klik Jul 3 '14 at 12:50
  • $\begingroup$ I will write an answer. In the future, please make your questions a little more clear. It makes it difficult for people who are willing to help you to do so. $\endgroup$ – Klik Jul 3 '14 at 13:10
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A 25.95g sample of methanol at 35.6°C is added to a 38.65g sample of ethanol at 24.7°C in a constant pressure calorimeter If the final temperature pf the combined liquids is 28.5°C and heat capacity of the calorimeter is 19.3J/°C determine the specific heat of methanol? The heat capacity of ethanol is 2.46J/(g⋅°C)

You should have included the heat capacitance of the calorimeter. Think about it, if the calorimeter was at 24.7°C before the methanol was added and then it rose 3.8°C along with the liquids, where did the energy come from to increase its temperature? Since you know that the heat capacitance of the calorimeter is 19.3J/°C then you can find out how much energy the calorimeter absorbed to increase in temperature by multiplying its heat capacitance by its temperature increase. I.e. $19.3J/°C \cdot 3.8°C = $ Joules absorbed by the calorimeter

Then we simply apply the heat lost = heat gained concept.

Heat lost = Heat gained

$25.95g \cdot (35.6°C-28.5°C) \cdot S = 38.65g \cdot (28.5°C - 24.7°C) \cdot 2.46 J/(g\cdot °C) + 19.3J/°C \cdot 3.8°C$

After solving for S we get $S = 2.36 J/(g\cdot°C)$

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  • $\begingroup$ Great. Glad I could help. $\endgroup$ – Klik Jul 12 '14 at 23:13

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