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Questions (7-9)
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In this lesson, questions (7-9) 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
C is Correct answer
Dheeraj Sharma
4 months ago
Absolutely Correct
in question no 8a in the calculation of entropy change why we not consider mass of copper
Suraj prasad singh
2 months ago
actually in the question it is given joule per kelvin not joule per kg kelvin
why not use integral form of entropy calculation to find the entropy change in lake?
Rafiq n
14 days ago
river acts like a reservoir... temp will never change
  1. COURSE: Detailed Explanation on Entropy Questions Lesson : Questions - (7-9)


  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 7 Ten grammes of water at 20 C is converted into ice at -10 C at constant atmospheric pressure. Assuming the specific heat of liquid water to remain constant at 4.2 J/gK and that of ice to be half of this value, and taking the latent heat of fusion of ice at 0 C to be 335 J/g, calculate the total entropy change of the system.


  4. Solution 7 mcp dT 293 273 293 0.01 4.2 In-_ kJ1K -0.00297 kJ/K 293 K --2.9694 J/K 2 273 K- -0.01 x 335 x 1000 273 -12.271 J/K 268 K 268 273 4.2 273 -0.3882 J/K . S4 Si 15.63 J/K Net Entropy change 15.63 J/K


  5. Question 8 Calculate the entropy change of the universe as a result of the following processes: (a) A copper block of 600 g mass and with Cp of 150 J/K at 100 C is placed in a lake at 8 C. (b) The same block, at 8 C, is dropped from a height of 100 m into the lake. (c) Two such blocks, at 100 and 0 C, are joined together.


  6. Solution 8 (AS) copper0 28 T (AS) coppermcp As the work is converted to heat and absorbed by water then (AS) iakeW (AS) univ = 0 + 2.09466 J/k = 2.09466 J/K 1-588-b J/ K = 2.09466 J/K 281 373 281 373 -42.48 J/K 281 (c) Final temperature T)1050 C 323 K As unit of Cp is J/K there for It is heat capacity C,-mc i.e. (ASCdT C,(100-8) 281 150(100-8) 281 (AS) lake J/ K (AS) system-150 In | +150 In- 150 3723-3.858 . J/ K 3.638 J/K (AS) univ- (AS) coP (AS)lake6.63 J/K 373 273 (b) work when it touch water = 0.600 x 9.81 100 J = 588.6 J As work dissipated from the copper


  7. Question 9 A system maintained at constant volume is initially at temperature T, and a heat reservoir at the lower temperature To is available. Show that the maximum work recoverable as the system is cooled to To is


  8. Solution 9 For maximum work obtainable the process should be reversible Cv = mcv Q- W (AS)rewoir (AS)eyele 0 H.E (Q1 - W) 20 Maximum work wmax-C, (T-T,) +T, Inl


  9. THANK YOU