## Homework Problem 8.21 in 6th edition: Q=mcat

$\Delta U=q+w$

Maayan Epstein 14B
Posts: 39
Joined: Fri Sep 28, 2018 12:20 am

### Homework Problem 8.21 in 6th edition: Q=mcat

Homework problem 8.23 in the 6th edition textbook states: "A calorimeter was calibrated with an electric heater, which supplied 22.5 kJ of energy as heat to the calorimeter and increased the temperature of the calorimeter and its water bath from 22.45 C to 23.97 C. What is the heat capacity of the calorimeter?"

I know that I need to use Q=mc(Tf-Ti) here, but I am not given the mass of either the calorimeter or the water.
The solutions manual solves for c this way: c=(Q)/(Tf-Ti)=22.5kJ (given)/(1.52 degrees C). This solution does not include mass, which confuses me because the Q=mc(delta T) equation does include mass. How are we able to solve this by disregarding m? Thanks!!

Kim Tran 1J
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

### Re: Homework Problem 8.21 in 6th edition: Q=mcat

I think the textbook used the equations: -q=qcal and Ccal= qcal/change in T
Ccal is the heat capacity of the calorimeter

q is the change in internal energy. So you can determine qcal is 22.5 kJ

then you can plug this into Ccal=qcal/change in T

change in temp= T2 - T1 = 23.97 C - 22.45 C = 1.52 C

Ccal = (22.5 kJ)/(1.52 C) = 14.8 kJ.C^-1