Answer to Question #194087 in Chemistry for kinzang

Question #194087

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?




1
Expert's answer
2021-05-17T04:09:07-0400

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.

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