[libra78wolf]

I recently conducted a experiment to determine the heat lost and heat gained of water and the heat capacity of calorimetry.

"The amount of heat gained by  the calorimeter is exactly equal to that lost by the hot water. The heat gained by the calorimeter may be expressed as the sum of two parts, where the first part represents the heat gained by the cold water in the calorimeter and the second part represents the heat gained by the cups plus the thermometer. show the calculation of the heat lost by the hot water and the heat gained. "

Okay so from my data I got I inserted it into the calculation
q=mass of cold water x specific heat x change in temperature of cold water. and the same for hot water.
Tfinal was 25.7 and my tinitial was 20.2 for cold water
Tfinal was 25.7 and my tinitial was 60.5.

q=39.420 x 4.179 x 5.5 = 906.04 cold water
q=47.431 x 4.179 x -34.8 = 6897.85 hot water

I don't understand if they are supposed to be the same according to my lab book, why are mine so far apart.
In desperate need of help

[savoy7]

libra -

some information we need to know -

in the setup of the lab - what was added to what?-

did you take hot water and add it to cold water? vice versa?

What absorbed q and what "lost" q?

It's difficult to help you with this problem when we don't know the experimental setup.

[libra78wolf]

hot water was added to cold

how much heat was gained by the cold and how much heat was lost by the hot water

[Demotivator]

First off, nice name. I can see how unbalaced equations/scales would drive a libra to desparation!  

true, Heat lost = heat gained. However, as stated, part of the heat gained is captured by the cups and thermometer which you have omitted (note, the problem includes the determination of calorimeter heat capacity, Ccal).
So,  6897.85 = 906.04 + Ccal(5.5)

[libra78wolf]

First of thank you but I screwed up. I need help with determining the temperature change now, before I can go ahead and determine the heat lost and gained.

My lab book:
Now you can calculate the change of temperature for the calorimeter and its cold water contents and hot water. To calculate the change in temperature use the following equation. ChangeT=Tfinal-Tinitial, where Tfinal is the extrapolated maxium time at the time of mixing and Tinitial is the temp[erature of the cold or hot water just before mixing. Showing the calculation of change in temperature cold water and change temperature hot water.

First off, am I looking for two changes in water temperature, one for the hot and one for the cold.
Secondly. I did my graph and the temperature at the time of mixing time 0 is 22.9. Would that be my Tfinal, or would the yintercept on my graph for the hot water be my extrapolated maxium temperature at the time of mixing.

My data
Time, minute  Temperature, °C
-4        20.0
-3        19.9
-2        20.0
-1        20.2
0        22.9
5.5        38.0
6        39.2
6.5        39.6
7        39.0
7.5        39.1
8        38.9
8.5        39.0
9        38.7
9.5        38.1
10        37.1

[Demotivator]

1) yes, change for hot and change for cold.
2) Tfinal is the extrapolated max temp to y intercept, not 22.9.

[libra78wolf]

So Change in Temeprature cold would be
40.5-20=20.5
and Change in Temperature hot would be
40.5-60.5=-20

My initial temp for cold was 20 and my initial temp for hot was 60.5.

40.5 being the y-intercept for hot water.