Question

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 1

508 J/kg/C

Step-by-step 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)