Chemistry help......

Question

A 100.8 g sample of a substance is initially at 22.0 °C . After absorbing 379 cal of heat, the temperature of the substance increases to 61.8 °C . What is the specific heat (𝑆𝐻) of the substance?

Answer 1

q = mC∆T

q = heat = 379 cal

m = mass = 100.8 g

C = specific heat = ?

∆T = change in temperature = 61.8º - 22.0º = 39.8º

Solving for C, we have...

C = q / (m)(∆T) = 379 cal / (100.8 g)(39.8º)

C = 0.0945 cal/gº (to 3 significant figures)