A 37.14 g sample of a substance is initially at 23.0 °C. After absorbing 1479 J of heat, the temperature of the substance is 125.6 °C. What is the specific heat ( 𝑐 ) of the substance?

Answer 1

q = mC∆T

q = heat = 1479 J

m = mass = 37.14 g

C = specific heat = ?

∆T = change in temperature = 125.6º - 23.0º = 102.6º

1479 J = (37.14 g)(C)(102.6º)

C = 0.388 J/gº