## 7.21 Set up of Calorimeter Problem

$\Delta U=q+w$

Hannah Pablo 2H
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Joined: Fri Sep 26, 2014 2:02 pm

### 7.21 Set up of Calorimeter Problem

Problem:
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?

How do you set up the equation for this problem?

Denise 3L
Posts: 13
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.21 Set up of Calorimeter Problem

You want to use the equation highlighted in section 7.5 (q=CdeltaT). You already have the given heat, and the change in temperature (in this case you don't need to convert it to Kelvin) and then solve for C, which will be your heat capacity. Since this is a calorimetry problem, you won't need q=nCdeltaT or q=gCdeltaT since you are solving for the heat capacity of the *calorimeter

Chem_Mod
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### Re: 7.21 Set up of Calorimeter Problem

Since the mass of the water bath was not given, you can assume that the water bath is a part of the calorimeter too. In this case, the whole system absorbs heat and increases in temperature. There's nothing more to do than find the ratio of these two quantities.

$C = q / \Delta T$

Justin Le 2I
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.21 Set up of Calorimeter Problem

For clarification, you don't need to convert Celcius into Kelvin because the size of the two units of temperature is the same.

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