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A power cycle operating between hot and cold reservoirs at 500 K and 300 K, respectively, receives 1000 kJ by heat transfer from the hot reservoir. The magnitude of the energy discharged by heat transfer to the cold reservoir must satisfy

User Ludington
by
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1 Answer

1 vote

Answer:

The value is
Q_l \ge 600\ k J

Step-by-step explanation:

From the question we are told that

The temperature of hot is
T_h = 500 \ K

The temperature of cold is
T_c = 300 \ K

The energy received is
E = 1000 \ kJ = 1000 *10^(3 ) \ J

Generally the maximum thermal efficiency of the engine is mathematically represented as


\eta = (T_h - T_c)/(T_h)

=>
\eta = (500 - 300)/(500)

=>
\eta = 0.4

Generally the thermal efficiency of the engine is


\eta_t = (Q - Q_l)/(Q)

Here
Q_l is the heat energy rejected

Generally the thermal efficiency must be less than or equal to the maximum thermal efficiency

So


(Q - Q_l)/(Q) \le 0.4

=>
(1000 *10^(3) - Q_l)/(1000 *10^(3) ) \le 0.4

the change in inequality sign is because
1000*10^(3) which was dividing started multiplying

=>
Q_l \ge 1000*10^(3) - 400*10^(-3)

=>
Q_l \ge 600*10^(3) \ J

=>
Q_l \ge 600\ k J

User SimonRH
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5.2k points