A boiling liquid absorbs thermal energy (heat) at a rate of 450 W. The specific latent heat of vaporisation is 2.7 × (10 race to power 6)J / kg. how much liquid is vaporised in 9 minutes?

0.09 kg

Explanation:

The details of boiling liquid are;

The rate of thermal energy absorption by the liquid, P = 450 W

The specific latent heat of vaporization, c = 2.7 × 10⁶ J/kg

The required information = The amount of liquid vaporized in t = 9 minutes

We have;

P = E/t

Where;

P = Power supplied by the heat source = The rate of absorption of thermal energy

E = The energy absorbed

t = The time

The energy absorbed in 9 minutes, E = P × t

∴ E = 450 W × 9 min × 60 s/min = 243,000 J

The heat required to evaporate 1 kg of liquid = The  specific latent heat of vaporization, c = 2.7 × 10⁶ Joules

The mass of the liquid evaporated by the energy absorbed, E, m = E/c

∴ m = 243,000 J/(2.7 × 10⁶ J/kg) = 0.09 kg

Therefore, the mass of the liquid vaporized by the energy absorbed at a rate of 450 W, in 9 minutes, m = 0.09 kg.

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