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Questions (10-12)
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In this lesson, questions (10-12) have been explained.

Harshit Aggarwal
Cleared UPSC ESE twice with Rank 63 and 90 in mechanical engg. Got 99 percentile in GATE. Cracked ONGC, BHEL,ISRO, SAIL, GAIL successfully

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  1. COURSE: Detailed Explanation on Entropy Questions Lesson : Questions -(10-12)


  2. About me Graduated from NIT Nagpur in 2008 * Cleared Engineering Services Examination (ESE-UPSC) Exam Got the offer letter from most of the Maharatna and Navratna Companies , Cleared GATE Exam * Rate, Review, Recommend, Share Follow me on Unacademy at: https://unacademy.in/user/harshit aggarwal


  3. Question 10 A body of finite mass is originally at temperature Ti, which is higher than that of a reservoir at temperature T2. Suppose an engine operates in a cycle between the body and the reservoir until it lowers the temperature of the body from T to T2, thus extracting heat Q from the body. If the engine does work W, then it will reject heat Q-W to the reservoir at T2. Applying the entropy principle, prove that the maximum work obtainable from the engine is W (max)Q- T2 (S S2) Where S - S2 is the entropy decrease of the body If the body is maintained at constant volume having constant volume heat capacity Cv = 8.4 kJ/K which is independent of temperature, and if Ti = 373 K and T2 = 303 K, determine the maximum work obtainable.


  4. Solution 10 Final temperature of the body will be T2 Wmax = [Q-T2 (Sl-S2)] wmax = Q-T2 (S1-S2) (AS) reservoirQ-W .. (AS) H.E.0 =Cv (T1-T2) + T2 Cv In Q- W (AS) univ. = (S2-S, ) +--20 303 273 = 8.41 373-303 + 303 In or T2 (S2- Si) Q-W20 = 58.99 kJ or WsQ T2 (Si - S2)]


  5. Question 11 Each of three identical bodies satisfies the equation U CT, where C is the heat capacity of each of the bodies. Their initial temperatures are 200 K, 250 K, and 540 K. If C= 8.4 kJ/K, what is the maximum amount of work that can be extracted in a process in which these bodies are brought sto a final common temperature?


  6. Solution 11 U=CT Therefore heat capacity of the body is C Let find temperature will be (T) For minimum T 8.4 kJ/K 540 K = 540 250 200 Q-Q+Q .. T300 K H.E (AS) 540K body ClnkJ/ K 540 Qi-W Q2 . 8.4(540 300) 2016 kJ (AS) 250K CIn Qi-wi = 8.4(300-250) = 420 kJ Q2-We = 8.4(300-200) = 840 kJ 250 250 K 200 K (AS) (AS) surrounds 0 . ( . Q1 Q2 (WW2) 1260 200K = C In 200 (Wi + w2) = 2016-1260 kJ = 756 kJ Wmax 756 kJ (AS)H.E0 or (AS)unit.- C I 540x 250 x 200


  7. Question 12 In the temperature range between 0 C and 100 C a particular system maintained at constant volume has a heat capacity CU = A + 2BT With A heat reservoir at 0 C and a reversible work source are available. What is the maximum amount of work that can be transferred to the reversible work source as the system is cooled from 100 C to the temperature of the reservoir? A = 0.014 J/K and B = 4.2 10-4 J/K?


  8. Solution 12 Find temperature of body is 273 K 373 K (AS)res.- 2W (AS)surrounds 0 273 373 Q- W 273 (AS)univ-,-0.08837 + -2 0 -A(100 B(2732 3732)J I.F or24.125Q W20 or W sQ- 24.125 or W s (28.532 -24.125) J or Ws 4.407 J --28.532 J (low from the system) 273 K (AS) boly C 373 A2BT dT Wmax 4.407 J A 2B (273-373) JI K -0.08837 J/K 273 373


  9. THANK YOU