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

Answer:

1356 s

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:

$P=\frac{E}{t}$

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

$t=\frac{E}{P}=\frac{3.39\cdot 10^6}{2500}=1356 s$