A 50.16 g sample of a substance is initially at 23.2 °C. After absorbing 2729 J of heat, the temperature of the substance is 196.3 °C. What is the specific heat (SH) of the substance?

Answer 1

q = m c ΔT

q = 2729 J

m = 50.16 g

ΔT = T_f - T_i = 196.3 - 23.2 = 173.1 °C

Rearrange and solve for c (specific heat)

c = q / m ΔT = 2729 J / (50.16 g x 173.1 °C) = 0.3143 J g^-1 °C^-1