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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

Unacademy user
nice lecture... can u please sir make a video dat summarize all the imp formulae at the end of chapters.
  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: 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