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

q= mC∆T is the equation to use.

q = heat = 1143 J

m = mass = 57.73 g

C = specific heat = ?

∆T = change in temperature = 169.6º - 20.0º = 149.6º

Solving for C we have..

C = q/(m)(∆T) = 1143 J/(57.73 g)(149.6º)

C = 0.132 J/gº