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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

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Mr. Roman you must do some changes in this app . The previous app version is best. Even mobile data is able to View lessons in previous version of this app. Your efforts are so good .No doubt about it but This new version app is Worst . Plz make some changes like previous version
rather than just reading out sir just explain the concept and skip the slides
1. COURSE : Detailed Explanation on Entropy Questions Lesson : Questions - (22-24)

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 22 An insulated 0.75 kg copper calorimeter can containing 0.2 kg water is in equilibrium at a temperature of 20 C. An experimenter now places 0.05 kg of ice at 0 C in the calorimeter and encloses the latter with a heat insulating shield. (a) When all the ice has melted and equilibrium has been reached, what will be the temperature of water and the can? The specific heat of copper is 0.418 kJ/kg K and the latent heat of fusion of ice is 333 kJ/kg. (b) Compute the entropy increase of the universe resulting from the procesS. (c) What will be the minimum work needed by a stirrer to bring back the temperature of water to 20 C?

4. Solution 22 Mass of ice 0.05 kg (a) Let final temperature be (T) 0.75 0.418% (293-T ) + 0.2 4.187 (293-T ) 333 x 0.05 +0.05 x 4.187 or 1.1509(293-T) or 337.2137 1.1509 T x (T 273) w,,-0.2 kg C,20.75 kJ /kg-K T = 293 K = 16.65-57.15255 + 0.20935 T or 277.68 K 4.68" C (b) (AS)aystem :0.75 x 0.4 18 In] oon | + 0.2 x 4.187x ln| 293 T, 273 293 333 0.05 273 - 0.00275 kJ/K 2.75 J/K +0.05 x 4,187 n (c) Work fully converted to heat so no Rejection = C (20-4.68) = 20.84 kJ C (Heat capacity) 1.36025

5. Question 23 Show that if two bodies of thermal capacities Ci and C2 at temperatures T and T2 are brought to the same temperature T by means of a reversible heat engine, then C,+C

6. Solution 23 dT H.E (AS)univ- (AS)| + (AS)2 (Ci + C2) In T = C1 In Ti + C2 In T2 Q- W For reversible process for an isolated system (AS) since In T - ClnT +C.ln T, n 2 Proved 0=C1 In T C

7. Question 24 Two blocks of metal, each having a mass of 10 kg and a specific heat of 0.4 kJ/kg K, are at a temperature of 40 C. A reversible refrigerator receives heat from one block and rejects heat to the other. Calculate the work required to cause a temperature difference of 100 C between the two blocks.

8. Solution 24 Mass 10 kg C 0.4 kJ/kg -K T 40 C 313 K (AS) ot- me In 313 T,-100 313 (AS) cold mcln For minimum work requirement process must be reversible so (AS)univ0 T1 313 K Q + W TT)o In 1 (313) T-100 T,-3132 :0 or T1 313 K Tf- 100 100 1002 4 x 313 or T 367 K or (-267) . Q+ W 10x 10.4 x (367-313) 215.87 kJ Q 10 x 0.4 x (313- 267)- 184 kJ Wmin 31.87 kJ

9. THANK YOU