a swimmer coming out from pool is covered with a film of water weighting about 18gm. how much heat must be supplied to evaporate is water at 298 k?
 

Asked by snigdha goel | 20th Sep, 2014, 06:12: PM

Expert Answer:

Dear ajaygoel0702@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.

The total energy is sum of two steps:

Heat from 25 °C to 100°C + Change to vapour

 First step: Heat from 25 °C to 100°C

Q1 = m x c x ΔT

Q1 = 18 x 4.18 (Sp. Heat of water) x (100 – 25)

    = 5643 J = 5.64 kJ

 Second step: Vaporization

The heat of vaporisation is 40.6 kJ mol-1 for water. Convert 18g of water to moles i.e. 1 mole.

Q2 = 40.6 x 1 = 40.6 kJ

 

The total energy = 40.6 + 5.64 = 46.24 kJ



Regards

Topperlearning Team.

Answered by Arvind Diwale | 22nd Sep, 2014, 02:57: PM