Imat 2020 ques 58

A beaker contains 1000g of a liquid that is stirred at its boiling point. A 100W electric heater is
completely immersed in the liquid. The heater provides the liquid with thermal energy, and 200g
of the liquid changes to vapour in 1600s.
What is the specific latent heat of evaporation of the liquid?
[Assume that no thermal energy is transferred to or from the surroundings, and no vapour
condenses.]
12.5J/g
160J/g
3200J/g
800J/g
16000J/ g

Hi!
Power = work/time

Work = power x time

Work = (100 W) x (1600 s)

Work = 160000 J

This work corresponds with the heat (Q) supplied to the liquid

Q = 160000 J

The specific latent heat (λ) of evaporation of the liquid is λ = Q/m

λ = (160000 J) / (200 g)

λ = 800 J/g.
Answer is thus D.
Hope it helps:)

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