A 53.93 g sample of a substance is initially at 21.3 °C. After absorbing 1993 J of heat, the temperature of the substance is 145.8 ° C. What is the specific heat ( c ) of the substance?

Answer 1

q = mC∆T

q = heat = 1993 J

m = mass = 53.93 g

C = specific heat = ?

∆T = change in temperature = 145.8 - 21.3 = 124.5º

1993 J = (53.93 g)(C)(124.5 deg)

C = 1993 J/(53.93/g)(124.5 deg)

C = 0.2968 J/g/deg