Question

A 40.32g sample of a substance is initally at 25.6 C. After absorbing 2625 J of heat, the temperature of the substance is 101.6 C. What is the specific heat (c) of the substance?

c= ________ J/g C

Answer 1

Q = mepat Q is the amount of heat required m is the mass of the substance Cp is the specific heat of the substance AT is the temperature difference AT = Toti Ta final temperature Tit initial temperature.

Q = mlp at Q = 26251 26251 = ( 40.329 ) (p) (101.6-956) 26257 = (40.329) (Cp) (26) p= 0.85663377193 3