A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific heat of the block?

Answer:
508 J/kg/C
Explanation:
Energy Lost by water = Energy gained by block
mcT = mcT [bolded is for water, italicised is for block]
(0.217)(4186)(25 - 16.4) = (0.350)(c)(16.4 + 27.5)
15.365c = 7811.9132
c = 508 J/kg/C (3 sf)