Question #26819

1 Answer
Al E.
Jul 15, 2017

#119.0 J/(°C)#, or #C.#

Explanation:

In a constant-volume calorimeter, we relate the heat capacity of the calorimeter to the temperature change in said calorimeter with the equation:

#q_(soln)=-C_(cal)*DeltaT#

The mass of the substance in the calorimeter in this case is unnecessary; some professors put these data in problems to confuse students who don't really know their stuff. Oh, test curves...

Anyways,

#-250.0=-C_(cal)*2.1°C#
#C_(cal)=1.2*10^2 J/(°C)#

Note, the question says released, thus the sign for the heat (#q#) is negative, not positive. Moreover, in my actual calculation, #C.# is the answer; here I have truncated for clarity.

Related questions
See all questions in Enthalpy
Impact of this question
1356 views around the world
You can reuse this answer
Creative Commons License