A 57.73 g sample of a substance is initially at 20.0 °C. After absorbing 1143 J of heat, the temperature of the substance is 169.6 °C. What is the specific heat ( 𝑐 ) of the substance?

Answer 1

For these types of questions, the equation Q = m • c • deltaT is used.

Hear Q is the enthalpy or heat transfer, m = mass, c = specific heat capacity and deltaT = change in temperature

In this case, we also convert 57.73 g to 0.05773 kg since c is written in J/kg•K in conventional unites.

1143 = 0.05773 • c • (169.6-20)

c = 132.3 J/ kg • K