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 1 of 2.
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)
Sunil kumar said:
1 decade ago
Process 1-2-3:
Qin-wout = u3-u1.
du = u3-u3 = 100-60 = 40.
Process 1-4-3:
Qin-wout = u3-u1.
Qin = wout+u3-u1.
Qin = 20+40 = 60.
Qin-wout = u3-u1.
du = u3-u3 = 100-60 = 40.
Process 1-4-3:
Qin-wout = u3-u1.
Qin = wout+u3-u1.
Qin = 20+40 = 60.
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)
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)
SANJEEV KUMAR said:
9 years ago
£(Q1 - Q2) = £(W1 - W2).
100- Q2 = 60 - 20,
Q2 = + 60 kJ.
100- Q2 = 60 - 20,
Q2 = + 60 kJ.
(1)
Ram said:
1 decade ago
q=u+w
1 case
100=u+60
u=40.
2 case
q=u+w
q=40+20
q=+60KJ.
1 case
100=u+60
u=40.
2 case
q=u+w
q=40+20
q=+60KJ.
Sanjay Patel said:
9 years ago
Very simple explanation, Thanks @Sanjeev Kumar.
Raghu said:
9 years ago
Thanks for your explanation @Ram & @Sunil.
Gopi said:
8 years ago
How to convert signs of heat& work?
(1)
Babu said:
9 years ago
Thanks for the solution @Ram & @Sunil.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers