Question

QUESTION 7 1.3 Calculate the intenal energy change in a process that performs 85.5 J of work on the system and produces 14.22 kJ of heat absorbed from the surround ings. OA 14.31 kJ OB.-14.13 kJ OC. 71.28 kJ OD.-14.31 kJ OE. 99.72 kJ OF. -99.72 kJ G. 14.13 kJ OH.-71.28 kJ

Answer 1

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According to first law of Thermodynamics Q = dutw Q = heat absorbed or released du= change in internd energy We work done by the system or on the syster Here, the heat is given to the system, from that some amount is used to change internal energy and the other is used to do some work. there, q= 14.22 KJ = 14220J of heat is absorbed (. positive) and w = 85.57 work done by the system on surrounding in ave sigor; 30 W= – 85.5J Then du = a-w. sign 9150 motion 14220 J-(-85.51) - 142205 +85.55 - 14305 5 J - 14.305 kJ 2 14.31 KJ Answer is 14.315