Question

Find the specific heat of an unknown metal given this data:
Mass of the metal = 120 g
Temp of the metal = 220 degrees
Mass of the water = 30 g
Initial Temp of the water = 20 degrees
Final Temp of the water = 85.6 degrees
Specific Heat of the water = 4.186 J/gram C

I get 0.51 J/gram C.

Answer 1

The sum of the heats gained is zero (one gains, the other loses).
Mm*Cm*(Tf-Tmi)+mw*Cw*(Tf-twi)=0

Cm= -mw*cw*(85.6-20)/(Mm(85.6-220)

= -30* 4.186*65.6/120*(-134.4)
= .51J/gC

Your answer looks good.