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Questions (25-27)
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In this lesson, questions (25-27) 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
Mam in second sentence can't we write "will they come in marriage?"
Hi, we can't use 'in marriage'. We will soon cover the lesson about use of prepositions so stay tuned for more lessons.
Monika Dahiya
2 years ago
Thanks mam
Sneha burman
10 months ago
mam I also face problem in preposition
thank you sir , it help me a lot
  1. COURSE : Detailed Explanation on Entropy Questions Lesson : Questions - (25-27)


  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 25 A block of iron weighing 100 kg and having a temperature of 100 C is immersed in 50 kg of water at a temperature of 20 C. What will be the change of entropy of the combined system of iron and water? Specific heats of iron and water are 0.45 and 4.18 kJ/kg K respectively


  4. Solution 25 Let final temperature is tr C 100 0.45 (100-tr) = 50 4.18 (tr-20) 100 t4.644 tr- 20 X 4.699 or 5.644 tf 192.88 tf 34.17320 C tr 307.1732 K or ENTROPY = 100 0.45 ln(3071732) = 1.1355 kJ/K (AS)iron0772 .50x 4.180 x ln 307.1732 \293 373


  5. Question 26 36 g of water at 30 C are converted into steam at 250 C at constant atmospheric pressure. The specific heat of water is assumed constant at 4.2 J/g K and the latent heat of vaporization at 100 C is 2260 J/g. For water vapour, assume pV= mRT where R = 0.4619 kJ/kg K, and -P = a + bT+ cP, where a = 3.634 b = 1.195 10-3 K-1 and c = 0.135 10-6 K-2 Calculate the entropy change of the system.


  6. Solution 26 m = 36 g 0.036 kg T1 = 30 C 303 K T2 = 373 K Ts 523 K (AS) Water 373 kJI K 0.03143 kJ/K CT2 mL (AS) Vaporization 523 373 0.036 x 2260 373 0.21812 kJ/K 523 dT mRla bx (523 373)(5232 - 3732) 2 0.023556 kJ/kg (AS) vapor mcPT (AS) System (AS) water (AS) vaporization + (AS) vapor 273.1 J/FK 373 523 373


  7. Question 27 An ideal gas is compressed reversibly and adiabatically from state a to state b. It is then heated reversibly at constant volume to state c. After expanding reversibly and adiabatically to state d such that Tb - Ta, the gas is again reversibly heated at constant pressure to state e such that Te - Tc. Heat is then rejected reversibly from the gas at constant volume till it returns to state a. Express Ta in terms of Tb an d Tc. If 1b 555 K and Te 835 K, estimate T. Take = 1.4.


  8. Solution 27 (AS) de = C, ln- or V C n-a Given Th = 555 K, Te = 835 K, = 1.4 + Gas (555)1.4+1 = 313.286 K 83514 (AS) cycles 0


  9. THANK YOU