Question

If 5.6 g of aluminum at 65.2 °C is placed in a calorimeter that contains water at 23.5 °C and the temperature of the water after the heat exchange occurs is 24.7 °C, then identify how much water is in the calorimeter.

I know this has something to do with q=mc delta T but I don't know what!!

So change in temperature is 1.2 C and specific heat is 4.18 j/g C but I don't know how to find Q. If I could find Q then I would just use m=q/c*delta T.... idk?

Answer 1

I don't get it. You have it worked out in your mind. Heat lost by Al is q = mcdT. Heat gained by water is q = mcdT.
heat lost by Al + heat gained by H2O = 0 so you never need to find q by itself.
[5.6 x specific heat Al x (Tf-Ti)] + [mass H2O x specific heat H2O x (Tf-Ti) = 0
Tf Al is 24.7
Ti Al is 65.2
Ti H2O is 223.5
Tf H2O is 24.7

The only unknown is mass H2O