Question

A sample of an unknown substance has a mass of 0.465 kg. If 3,000.0 J of heat is required to heat the substance from 50.0°C to 100.0°C, what is the specific heat of the substance?
Use q equals m C subscript p Delta T..
0.00775 J/(gi°C)
0.0600 J/(gi°C)
0.129 J/(gi°C)
0.155 J/(gi°C)

Answer 1

ΔT = 100.0 − 50.0 = 50.0°C. Convert mass: 0.465 kg = 465 g.

c = q / (m ΔT) = 3000 J / (465 g × 50.0°C) = 3000 / 23250 = 0.129 J/(g·°C).

Answer: 0.129 J/(g°C).