Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • Can someone explain . A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 105 Nm-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of th

. A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 105 Nm-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

  • Option 1)

    42.2 J

  • Option 2)

    208.7 J

  • Option 3)

    104.3 J

  • Option 4)

    84.5 J

 

Answers (2)

best_answer

As we have learned

 

First law of Thermodynamics -

Heat imported to a body in is in general used to increase internal energy and work done against external pressure.

- wherein

dQ= dU+dW

 

 

Heat\: Supplied= 54 cal = 54\times 4.2\: J= 226.8\: J

Work\: Done= 54 cal =P\Delta V\: = 1.01\times 10^{5}\times 167.1\times 10^{-6}J

                            = 168.8\times 10^{-\gamma }= 16.88\: J

\therefore \: \: \: \Delta U= 226.8-16.88J\simeq 208.7J

 


Option 1)

42.2 J

This is incorrect

Option 2)

208.7 J

This is correct

Option 3)

104.3 J

This is incorrect

Option 4)

84.5 J

This is incorrect

Posted by

Plabita

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NTA re-opens NEET UG correction window 2024; update photographs till April 26
anam-khan

NEET UG 2024 exam city slip out at neet.ntaonline.in; over 25 lakh candidates to appear
anam-khan

NEET UG city intimation slip likely today at exams.nta.ac.in/NEET; barred item at exam centre

Latest Question