A 59.02 g sample of a substance is initially at 29.6 ∘C . After absorbing 244 cal of heat, the temperature of the substance is 164.7 ∘C . What is the specific heat (𝑆𝐻) of the substance?

Answer 1

q = mC∆T

q = heat = 244 cal

m = mass = 59.02 g

C = specific heat = ?

∆T = change in temperature =164.7 - 29.6 = 135.1º

Solving for C, we have C = q/m∆T

C = 244 cal/(59.02 g)(135.1º)

C = 0.0306 cal/gº