A 29.85 g sample of a substance is initially at 25.2 °C. After absorbing 2189 J of heat, the temperature of the substance is 119.9 °C. What is the specific heat (c) of the substance?

Answer 1

q = m c ΔT

where q = the amount of heat absorbed or released

m = mass

c = specific heat

ΔT = change in temperature = T_final - T_initial

rearrange the equation to solve for c, you get

c = q / m ΔT

q = + 2189 J (q is positive since heat was absorbed by the substance)

m = 29.85 g

T_i = 25.2 °C

T_f = 119.9 °C

c = 2189 J / (29.85 g x (119.9 - 25.2 °C))

= 0.774 J / g °C