A 44.69 g sample of a substance is initially at 21.0 °C. After absorbing 2901 J of heat, the temperature of the substance is 120.4 °C. What is the specific heat ( 𝑐 ) of the substance?

Answer 1

q = mC∆T

q = heat = 2901 J

m = mass = 44.69 g

C = specific heat = ?

∆T = change in temperature = 120.4º - 21.0º = 99.4º

Solving for C:

C = q / (m)(∆T)

C = 2901 J / (44.69 g)(99.4º)

C = 0.653 J/gº