Determine the specific heat of copper from the fact that 38 g of 80°C copper are needed to raise the temperature of 15 g of water from 22°C to 33°C

From the equation q=mCΔT, set the q of copper = to q of water,

So - - - mCΔT (copper) = mCΔT (water).

mass (Cu - copper) = 38g

mass (H2O - water) = 15g

C (H2O) = 4.184 J/g*C

ΔΤ (H2O) = 33-22 = 11*C

ΔΤ (Cu) = 33-80 = - 47*C (the final temp is the same for both materials - thermal equilibrium)

C (Cu) = ?

So - - - 38 (-47) C[Cu]=15 (4.184) (11)

- - - C[Cu]=690.36 / (-1786) = 0.3865 J/g*C, or 0.39 in 2 sig figs. (The negative goes away, because specific heats are usually positive)