Question

Assume you are performing the calibration step of Experiment 8 and you begin with 20 g of water at 20 oC and 20 g of water at 80 oC. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 oC. What is the heat capacity of the calorimeter? Assume room temperature is 25 oC.

Answer 1

quation to use:
-q (warm water) = q (cool water) + q (calorimeter)

which then turns into this equation.
- specific heat (warm water) X mass (warm water) X delta Temp =
specific heat ( cool water ) X mass ( cool water ) X delta Temp + C (calorimeter) X delta Temp

Now here comes the actual calculations

I first start by subtraction q (cool water) over the equal sign which will give me
-q (warm water) - q (cool water) = q (calorimeter)

Time to input our given values

specific heat of water = 4.184 (J / grams) * C ( Celsius )
*Note other people tried to solve this problem by trying to calculate the specific heat of water at the temperatures that were given in the problem. Do not do this.

C (calorimeter) X ( 45 - 25 ) =
-(4.184 (J / grams) * C) X (20 grams) X (45 - 20) - (4.184 (J / grams) * C) X (20 grams) X (45 - 80)

*The first change of temperature (45 - 25) is for the calorimeter. It basically means that the calorimeter starts at 25 C ( Celsius ) (because of room temperature) and ends at 45 C ( Celsius ) after the warm and cold mixture.

Once these values are calculated you should get this.
C (calorimeter) X ( 20 C ( Celsius ) ) = -2092 ( cool water ) + 2928.8 ( warm water )

Which will then form this.
C (calorimeter) X ( 20 C ( Celsius ) ) = 836.8
Now simply divide 836.8 by 20 and you should get.

C (calorimeter) = 41.84 J / C ( Celsius )

Now to convert to cal / C ( Celsius )

41.84 J / C ( Celsius ) X ( 1 cal / 4.18400 J ) = 10 cal / C ( Celsius )