Temperature of colder water : 23C
Temperature of warmer water : 66C
Final temp of water after mixing : 42C
Heat lost by the warmer water : -5020.8J ,(209.2)(42-66)
Heat gained by colder water : 3974.8J ,(209.2)(42-23)
Heat absorbed by the calorimeter:??
Heat Capacity, C, of calorimeter :??

I really really cannot find these two answers...
I think this experimental was easy but calculation parts are so hard and confusing.

Please Please help me.. I really cannot do these two questions..

Thanks

From the principle of conservation of energy, heat gain should equal to the magnitude of heat lost.

Heat gained = 3974.8 J
Heat lost = 5020.8 J

Since there is a difference, that difference has got into the calorimeter.

So, heat dissipated to calorimeter = 5020.8 - 3974.8 =1046 J

Now, heat capacity is given by:

C = \frac{Q}{\theta}

Q is the amount of energy
Theta is the change in temperature of the calorimeter.

Can you finish that now?

So heat absorbed by the calorimeter is 1046J
correct?
heat dissipated to calorimeter = 5020.8 - 3974.8 =1046 J
And how can I find Q is the amount of energy
Theta is the change in temperature of the calorimeter.

Thanks

Q here is exactly equal to 1046 J.

The energy dissipated from the mixture to the calorimeter is that amount of energy, assuming no other heat loss occurs.

The temperature change depends on your experiment. What was the initial temperature of the calorimeter? It's final temperature? Assuming that the temperature of the calorimeter is the same as that of the mixture, you can find it. :)

Help! I am trying to find the amount of enegy gained by my homemade calometer, but it just doesn't make sense. If the temperature changed from 2.6 to 3.6 that is one degree. It then says I should multiply that by the mass of the oil- 145 g for my answer, but I know the marshmellow only has 25 cals.
Things are not adding up, I must be doing something wrong.