A 2.05 g sample of a metal alloy wire is heated to 98.88°C. It is then quickly dropped into 28.0 g of water

A 2.05 g sample of a metal alloy wire is heated to 98.88°C. It is then quickly dropped into 28.0 g of water in a calorimeter. The water temperature rises from 19.73°C to 21.23°C. Calculate the specific heat of the alloy.

q=mcT
Energy lost by alloy=energy gained by water
Therefore must find energy gained by water (Specific heat capacity of water is 4.2J/gK):
Change in temperature is 1.5 degrees
q=28g x 4.2 x 1.5
q=176.4J therefore 176.4J lost by alloy
Alloy:
change in temperature is 77.65 degrees
c=q/mT
c=176.4/(77.65 x 2.05)
c=1.108J/gK

The heat gained by the water equals the opposite in magnitude of the heat lost by the alloy. Heat = Q = mass * heat capacity * temperature change

So for the water, Q = 28 * 2.184 * (21.23-19.73) = 91.728 J

So this -91.728 J = ,m(alloy) * specific heat(alloy) * delta T = 2.05 * C(alloy) * (21.23-98.88)

Specific Heat of Alloy = .576 J/deg.K