Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 6 (Q.No. 2)
2.
A path 1-2-3 is given. A system absorbs 100 kJ as heat and does 60 kJ of work while along the path 1-4-3, it does 20 kJ of work. The heat absorbed during the cycle 1-4-3 is
Discussion:
18 comments Page 2 of 2.
Narendra said:
8 years ago
Thank you for explaining it.
Gopi said:
8 years ago
How to convert signs of heat& work?
(1)
Mohammed saif said:
7 years ago
U1=U2 ==>Q1-W1=Q2-W2,
100-60=Q2-20,
Q2= 60 KJ.
100-60=Q2-20,
Q2= 60 KJ.
(1)
Priyanka said:
7 years ago
Good explanation @Ram.
(1)
Mritunjay said:
7 years ago
This is the first law of Thermo for process but you are using this for cycle. Is it possible?
(1)
Salman said:
7 years ago
Thanks for all the explanation.
(1)
Pramodkumar said:
6 years ago
Internal energy a property, so it does not depend on path, internal energy for both path will be same.
Q = U+W.
100 =U+60,
then U= 40.
Then other path Q = 40 + 20 = 60 kJ.
Q = U+W.
100 =U+60,
then U= 40.
Then other path Q = 40 + 20 = 60 kJ.
(12)
Raghu mb said:
5 years ago
Thanks @Ram.
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