COURSE : Detailed Explanation on Entropy Questions Lesson : Questions - (25-27)
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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
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
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.
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
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.
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
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