the temperature of a 62 g sample of material invreases from 21 C to 41 C when it absorbs 3000 J of energy as heat.

Question

What is the specific heat of this material (round your answers to the nearest thousandth)

Answer 1

q = mC∆T

q = heat = 3000 J

m = mass = 62 g

C = specific heat = ?

∆T = change in temperature = 41º - 21º = 20ºC

3000 J = (62 g)(C)(20 deg)

C = 3000 J/(62 g)(20 deg)

C = 2.419 J/g/deg