Question

What is the shortest time for which a 2.5kW kettle containing 1.5kg of boiling water can be left without boiling dry? (Take the specific latent heat of vaporization of water as 2.26x10^6 J/kg).

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Answer 1

1356 s

Step-by-step explanation:

The total energy that must be supplied to the water to completely boil it is:

`E= m \lambda`

where

m = 1.5 kg is the mass of the water

`\lambda=2.26\cdot 10^6 J/kg` is the specifi latent heat of vaportization of water

Substituting,

`E=(1.5)(2.26\cdot 10^6)=3.39\cdot 10^6 J`

The power of the kettle is

`P=2.5 kW = 2500 W`

And power is defined as the ratio between energy supplied and time:

`P=(E)/(t)`

So, the shortest time the kettle can be left to boil the water is:

`t=(E)/(P)=(3.39\cdot 10^6)/(2500)=1356 s`