6th edition Hw problem 8.23


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beckyolmedo1G
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Joined: Fri Sep 28, 2018 12:21 am

6th edition Hw problem 8.23

Postby beckyolmedo1G » Tue Feb 05, 2019 4:26 pm

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 degrees C to 23.97 degrees C. What is the heat capacity of the calorimeter?

What equation is used in order to find the heat capacity?

Katie Sy 1L
Posts: 57
Joined: Fri Sep 28, 2018 12:18 am

Re: 6th edition Hw problem 8.23

Postby Katie Sy 1L » Tue Feb 05, 2019 5:00 pm

you want to find Ccal so take 22.5 kj and divide that by the difference in temperature. your answer will be in kJ/C


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