The temperature of a 85.6 g piece of a material increases from 21.0 C to 35.00 when it absorbs 648 J of heat. What is the specific of this substance?

Answer 1

q = mC∆T

q = heat = 648 J

m = mass = 85.6 g

C = specific heat = ?

∆T = change in temperature = 35.00 - 21.0ºC = 14º

Solving for C, we have..

C = q / (m)(∆T) = 648 J / (85.6 g)(14º)

C = 0.541 J/gº (3 sig. figs.)