The increase in internal energy of 1 kg of water at `100^(@)` C when it is converted into steam at the same temperature and 1 atm (100 kPa) will be :

Question

The increase in internal energy of 1 kg of water at `100^(@)` C when it is converted into steam at the same temperature and 1 atm (100 kPa) will be :

The increase in internal energy of 1 kg of water at `100^(@)` C when it is converted into steam at the same temperature and 1 atm (100 kPa) will be :
[The density of water and steam are `1000 kg//m^(3) "and" 0.6 kg//m^(3)` respectively. The latent heat of vapourisation of water is `2.25xx10^(6) J//kg`.]
A. `2.08xx10^(6) J`
B. `4xx10^(7) J`
C. `3.27xx10^(8) J`
D. `5xx10^(9) J`

Monjur Alahi

9 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A
latent heat of vaporisation of water `=2.25xx10^(6)J//Kg`
`{:(DeltaH=2.25xx10^(6)J//Kg,rArr,"work done"= -P_(ext)(V_(2)-V_(1)),,),(DeltaH=2.25xx10^(6),rArr,DeltaH=DeltaU+P DeltaV,,):}`
(a) Now, volume of water `V=((m)/(d))=(1)/(100)M^(3)=1L`
(b) volume of steam `=(1000)/(0.6)=[1666.67-1]101.325`
`Delta U=22.5xx10^(5)-1.68xx10^(5)=20.8xx10^(5)=2.08xx10^(6)J`