a student conducted an experiment to determine the specific heat of an undefined metal

Question

metal mass-45g

initial temp of metal-100 C

mass of water in styrofoam cup- 50g

initial temp of water in styrofoam cup- 23 C

final temp of water in styrofoam cup- 25 C

final temp of metal- 25 C

Answer 1

q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temperature

The heat lost by the hot metal MUST equal the heat gained by the cooler water in the styrofoam cup

Heat lost by metal = q = (45 g)(C J/gº))(100º - 25º) = 3375C

Heat gained by water = q = (50 g)(4.184 J/gº)(25º - 23º) = 418

(45 g)(C J/gº))(75º) = (50 g)(4.184 J/gº)(2º)

3375 C = 418

C = 0.124 J/gº = specific heat of the metal