## 8.23

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Joined: Fri Sep 29, 2017 7:05 am

### 8.23

Can someone explain how to do #23 " a calorimeter was calibrated with an electric heater, which supplied 22.5kJ 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?" Thanks in advance!

Diego Zavala 2I
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### Re: 8.23

Divide 22,500J by Delta T (23.97-22.45)Celcius to get J/C, which is the heat capacity of the calorimeter

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Joined: Fri Sep 29, 2017 7:07 am
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### Re: 8.23

I'll try a bit little more specific if it helps. So we use the equation $q_{cal}=C\bigtriangleup T$ which is the equation for heat transfer for a calorimeter. So then we move around to equation to get $\frac{q_{cal}}{\bigtriangleup T}=C$ which is the heat capacity of a calorimeter. Lastly, just plug in the given variables and you should get 14802 J/C if I did my math correctly.