80.0g of propanone was heated from 25.50C to 63.40C. How much energy was transferred to the propanone during heating? Specific heat of propanone is 2.13Jg-1 oC-1?
q = m × C × ΔT
q = amount of heat energy gained or lost by substance (J)
m = mass of sample (g)
C = specific heat capacity of sample (J °C-1 g-1)
ΔT = change in temperature (°C)
m(propanone) = 80.0 g
C(propanone) = 2.13 J °C-1 g-1
ΔT = (63.40 - 25.50)°C = 37.9°C
q = (80.0 g) × (2.13 J °C-1 g-1) × (37.9°C) = 6458.16 J = 6.46 kJ
q = 6.46 kJ (if you do not take into account the phase transition 'liquid --> gas')
Answer: 6.46 kJ of energy was transferred to the propanone during heating.