Question

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

Answer 1

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)