Specific Heat capacity

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Fenared Ortiz 3K
Posts: 37
Joined: Wed Nov 16, 2016 3:03 am

Specific Heat capacity

Postby Fenared Ortiz 3K » Tue Feb 13, 2018 9:38 pm

A 245.7g sample of a metal at 75.2 degrees Celsius was placed in 115.43g of water at 22.6oC . The final temperature of the water was 34.6oC . Assuming no heat was lost to
the surroundings, calculate the specific heat of the metal.

Are we supposed to use q=nCd*delta T, and if so then how to we get to the final answer of 0.581 J/G*C

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Re: Specific Heat capacity

Postby Andrea Grigsby 1I » Tue Feb 13, 2018 10:12 pm

so use q=mCdelta T to calculate the heat produced by the water when the temperature is raised from 22.6 to 34.6
so it would look something like this:
115.43*4.184*(34.6-22.6) = 5795.51 J
and then for the metal
because Qsys=-Qsurr
245.7*C*(34.6-75.2)= -5795.51 J
C=-9975.42/-5795.51
=0.580979
=0.581 J/GC


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