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Questions (16-18)
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In this lesson, questions (16-18) 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
sir y dont u explain few things,,you just keep on reading the solution !
  1. COURSE : Detailed Explanation on Entropy Questions Lesson : Questions -(16-18)

  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 16 (a) An aluminium block (cp 400 J/kg K) with a mass of 5 kg is initially at 40 C in room air at 20 C. It is cooled reversibly by transferring heat to a completely reversible cyclic heat engine until the block reaches 20 C. The 20 C room air serves as a constant temperature sink for the engine. Compute (i) the change in entropy for the block, (ii) the change in entropy for the room air, (iii) the work done by the engine. (b) If the aluminium block is allowed to cool by natural convection to room air, compute (i) the change in entropy for the block, (ii) the change in entropy for the room air (iii) the net the change in entropy for the universe

  4. Solution 16 293 (AS)A dT 313 293 313 5 x 400 x In/ K-132.06 J/K 313 K)5 kg 293 And Q-m cp (313- 293)40000 J As heat is reversibly flow then (AS)A (AS) air0 132.06 136.52- or 293 W 1.306 kJ 293 K or (b) (AS) Same for reversible or irreversible-132.06 J/K 4000 (AS) air- - 136.52 J/K 293 (AS) air4.4587 J/K

  5. Question 17 Two bodies of equal heat capacities C and temperatures Ti and T2 form an adiabatically closed system. What will the final temperature be if one lets this system come to equilibrium (a) freely? (b) Reversibly? (c) What is the maximum work which can be obtained from this system?

  6. Solution 17 Freely T, =--2 2 T,JC (b) Reversible Let find temperature be T dT the (AS)hot- I C Q-W (AS)cold-JCC (AS)univ. = (AS)hot-(AS)cold = (AS)su

  7. Question 18 A resistor of 30 ohms is maintained at a constant temperature of 27 C while a current of 10 amperes is allowed to flow for 1 sec. Determine the entropy change of the resistor and the universe. If the resistor initially at 27 C is now insulated and the same current is passed for the same time, determine the entropy change of the resistor and the universe. The specific heat of the resistor is 0.9 kJ/kg K and the mass of the resistor is 10 g.

  8. Solution 18 heat As resistor is in steady state therefore no change in entropy. But the work is dissipated to the atmosphere So (AS)at Rt atm 10 x 30 x1 - 10 kJ/kg 300 1 If the resistor is insulated then no heat flow to surroundings So (AS) surrounding 0 And, Temperature of resistance (At) 10,x30x1 = 333.33 C 900 x 0.01 Final temperature (T) 633.33 K Initial temperature (To) 300 K (AS)- m o 633.33dT 300 ( - 633.33 300 = 0.01 0.9 In 6.725 J/K (AS)univ - (AS)rev6.725 J/K