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.
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
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