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.
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.
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.
Subha said:
10 years ago
Thanks for your explanation.
Piyush said:
9 years ago
Thank you @Ram & @Sunil.
Raghu said:
9 years ago
Thanks for your explanation @Ram & @Sunil.
Swadhin said:
9 years ago
Thanks @Ram & @Sunil.
Shekhar said:
9 years ago
Thanks for explaining the solution.
Babu said:
9 years ago
Thanks for the solution @Ram & @Sunil.
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)
Sanjay Patel said:
9 years ago
Very simple explanation, Thanks @Sanjeev Kumar.
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