Two Carnot heat engines operate in tandem as follows: engine A takes in 13.0 kJ

Question

Two Carnot heat engines operate in tandem as follows: engine A takes in 13.0 kJ per cycle from a heat reservoir at a temperature of 470 K. The heat rejected by engine A is received by engine B, which performs 4.3 kJ of net work per cycle. Engine B, in turn, rejects heat at a temperature of 320 K. The temperature at which engine A rejects heat to engine B, in SI units, is closest to:

The answer is 475.
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Explanation / Answer

.We have 2 Carnot engine and between Engine A to Engine B, there is heat transfer. THA = 470 K QHA = 13 kJ TCB = 320 K WB = 4.3 kJ Eff A = 1 – TCA/THA = 1 – TCA/470…..(1) Eff A = 1 – QCA/QHA = 1-QCA/13….(2) (2)-->(1) QCA/13 = TCA/470 TCA = 36.154 QCA………(3) Eff B = 1-TCB/THB = 1-320/THB………..(4) Eff B = W/QHB = 4.3/QHB………………(5) (4 --> (5) 4.3/QHB = 1-320/THB….(6) Let’s assume there is virtual engine between A and B Eff A-B = W/QCA, no work added Eff A-B = 0 / QCA Eff A-B = 1 – QHB/QCA = 0 QCA = QHB…………….(7) Eff A-B = 1-THB/TCA = 0 THB = TCA……. (8) (8) and (7) --> (6) 4.3/QCA = 1-320/TCA = TCA – 320/TCA TCA/QCA – 4.3 = TCA – 320….(9) (3)-->(9) 36.154 QCA/QCA -4.3 = TCA-320 TCA = 351.1854 K

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