70 kilojoules of heat were added to a 1.5−kg piece of wood whose temperature then rose from 20°C to 49°C. What was the specific heat of the wood? c = ? kJ kg·°C

Answer 1

c = Q / (mΔT)

Q = 70 kJ

ΔT = (49 - 20)ºC

m = 1.5 kg

Plug the numbers into the equation for c and get the answer in units of kJ / (kgºC).