Question

While water is boiling, it absorbs 2257 kJ of heat per kilogram of water. This takes place at a

constant temperature of 100°C. What is the entropy change of 10 kg of water while it boils? (Be
careful of your energy units).

Answer 1

The correct answer is "6.015×10⁴ J/K".

Step-by-step explanation:

The given values are:

Mass of water,

= 10 kg

Latent heat of boiling,

= 2257 kJ

Constant temperature,

= 100°C

The entropy change will be:

=
`(mass* latent \ heat \ of \ boiling)/(temperature)`

On substituting the values, we get

=
`(10* 2257* 1000)/((273+100)K)`

=
`(22,570,000)/(373K)`

=
`6.051* 10^4 \ J/K`