Question

how much heat is absorbed when 1 kg of liquid gets vaporized at its boiling point

Answer 1

22.5×10 5 j is absorbed

Answer 2

$\bold{22.6 \times 10^{5} \mathrm{J}}$ of heat is absorbed when 1 kg of liquid gets vaporized at its boiling point.

Explanation:

‘Heat of vaporization’ is the amount of ‘heat’ required to convert '1g' of mass of a liquid to vapor. Normal “boiling point of water” is $100^{\circ} \mathrm{C}$. The heat of vaporization at this temperature is 2260 J/g. It means 2260 J/g of heat is required to convert ‘1g of water’ to 1g of vapor at $100^{\circ} \mathrm{C}$. So to convert 1 kg of mass of water or 1000g of water we require $22.6 \times 10^{5} \mathrm{J}$ of heat.