120g of copper at 100 degrees celsius is added to 80g of water in a 60g calorimeter at 20 degrees. If there is no heat loss to the surroundings, find the final temperature of the system.

I could be wrong, but here goes.
(Heat lost by the 120g chunk of copper added) = (Heat gained by 80g water) + (heat gained by copper calorimeter)
It's a closed system, no heat is lost to the surroundings, so once you identify what is losing heat and how much it loses, that must equal the whatever the other objects in the system gain.
There's no change of states, so we don't need to worry about latent heat.
We don't know what temperature the chunk of copper will cool to, so we'll call it x. Therefore the heat change of the chunk is 100 - x.
We know that the temperature of the water and calorimeter must also rise to this temperature. Since x will have to be a higher temperature than 20 degrees Celsius, we can say the temperature change is x - 20.
Using the equation for specific heat capacity and subbing in what we know into my word equation at the top of my answer, we get:
(120)(390)(100 - x) = (x - 20)(4180)(80) + (x - 20)(390)(60)
The algebra should be okay from here on in, it's just creating the equation that's the problem.
I got x = 29.25 C.

Chris241197— 17/09/16