what is the answer?

Question

A 33.02 g sample of a substance is initially at 27.8 °C. After absorbing 1905 J of heat, the temperature of the substance is 107.8 °C. What is the specific heat (𝑐) of the substance?

𝑐=

Answer 1

q = mC∆T

q = heat = 1905 J

m = mass = 33.02 g

C = specific heat = ?

∆T = change in temperature = 107.8º - 27.8º = 80.0º

Solve for C

C = q / (m)(∆T) = 0.721 J/gº