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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
U
Unacademy user
Very helpful Specially in the last minute study !
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thankyou deepak for your valuable feedback
in question no 8a in the calculation of entropy change why we not consider mass of copper
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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?
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Rafiq n
9 months 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