Question

a chemist conducts a calorimetry experiment of an unknown metal. the mass of the sample is 24.4g. the chemist heats the metal to 100 degrees celsius, then places it in a coffee-cup calorimeter containing 108.5g of water. the temp of the water rises from 21.3 degrees celsius to 25.0 degrees celsius. how much heat was absorbed by the water? how much heat was released by the metal? what is the specific heat of the metal?

Answer 1

mass of metal = 24.4 g

temperature of metal = 100 oC

mass of water = 108.5 g

temperature = 21.3 oC

heat absorbed by water = (m Cp dT) water

                                     = 108.5 x 4.184 x (25 - 21.3)

heat absorbed by water   = 1679.7 J

heat released by metal = 1679.7 J

heat loss by metal = heat gain by water

(m Cp dT)metal = (m Cp dT) water

24.4 x Cp x (100 - 25) = 1679.7 J

Cp = 0.918

specific heat of metal = 0.918 J / g oC