A 34.07 g sample of a substance is initially at 27.4 ° C.

Question

A  sample of a substance is initially at  After absorbing  of heat, the temperature of the substance is  What is the specific heat () of the substance?

Answer 1

q = mC∆T

q = heat = 1375 J

m = mass = 34.07 g

C = specific heat = ?

∆T = change in temperature = 181 - 27.4 = 153.6ºC

C = q/m∆T = 1375 J/(34.07 g)(153.6ºC)

C = 0.263 J/g/ºC