Question 26819

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:

${q}_{s o \ln} = - {C}_{c a l} \cdot \Delta T$

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.