## chemistry problem on temp. / heat transfer stuff

question is
100.0 grams of copper metal initially at 100.0 degrees Celsius, is added to a calorimeter, containing?
250.0 grams of H20 at 15.0 degrees Celsius, if the specific heat of copper is .389 J/g Celsius, what is the final temp of the water and copper mixture?
You can use this equation most likely c=q/t
And solve by rearranging the equation for delta t:
so you have delta t = q/c

the top part of the right side should all have to do with heat transfer?
the bottom part would have to do with temperature change?

it might look something like this but this doesn't seem to be the correct method, it doesn't seem right?

.389 j/g = (100g)*(100C-15C)(250)
Do you think that would give the correct answer?
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 Admin As you were already told elsewhere, you should use q=mcΔT, and do the full heat balance.