Question

A 2kW kettle containing boiling water is placed on a balance. It is left there and continues to boil for 5 minutes. The balance reading changes by 0.2kg. What does this information give as a value for the specific latent heat of vaporisation of water

Answer 1

Given that,

Energy = 2 kW

Time = 5 minutes

Mass = 0.2 kg

We need to calculate the latent heat of vaporization of water

Using formula of energy

$Q=mL$

$L=\dfrac{Q}{m}$

$L=\dfrac{P\times t}{m}$

Where, L = latent heat

P = power

t = time

m = mass

Put the value into the formula

$L=\dfrac{2\times10^{3}\times5\times60}{0.2}$

$L=3\times10^{6}\ J/kg$

Hence, The latent heat of vaporization of water is $3\times10^{6}\ J/kg$