chemcalorimetry
posted by natash on .
If a 5.26g sample of copper at 258C is placed in 125mL of water at 21.0C, how hot will the water get? Assume no heat loss to the surroundings. The specific heat of copper is .385 J/g C

heat lost by Cu + heat gained by water = 0
mass x specific heat Cu x delta T = q Cu.
mass x specific heat H2O x delta T = q H2O.
delta T is TfinalTinitial. You are given the initial T for both Cu and water, Tf is the unknown for which you solve. You must look up the specific heats of Cu and H2O. 
I have assumed in the above that the final T of the water is NOT greater than 100. If it is another approach must be used.

I'm a bit confused so is this the right way to set up the problem
heat lost by Cu+heat gained by water=0
5.26g(.385)(Tf258C)+125g(4.18)(Tf21C)=0 then you solve for Tf? 
Yes, you have it set up correctly. Solve for Tfinal.