Answer to Question #194087 in Chemistry for kinzang

Question

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?

Answer 1

Solution:

q = m × C × ΔT

where:

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

Thus:

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.