A 2657 g sample of a substance is initially at 225 C After a

A 26.57 g sample of a substance is initially at 22.5 °C. After absorbing 1057 J of heat, the temperature of the substance is 127.0 °C. What is the specific heat (c) of the substance?

Solution

by definition

q = mass x specific heat x temp change

q = m x c x dT

c = q / ( m x dT)

given

q = 1057 J

mass (m) = 26.57 g

dT = Tf - Ti = 127 C - 22.5 C = 104.5 C

plug in the values to get c

c = 1057 J / ( 26.57 g x 104.5 C)

c = 0.3807 J/ gC

so

the specific heat of the substance is 0.3807 J / g C