A thermodynamic cycle takes in heat energy at | vedclass

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(A) Let the heat absorbed be $Q_{1}$,the heat rejected be $Q_{2}$,and the work done be $W$.According to the first law of thermodynamics for a cycle,$Q

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$0.25$

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(A) Let the heat absorbed be $Q_{1}$,the heat rejected be $Q_{2}$,and the work done be $W$.According to the first law of thermodynamics for a cycle,$Q_{1} = W + Q_{2}$.Given that the energy rejected at the low temperature is $3$ times the work done,we have $Q_{2} = 3W$.Substituting this into the energy balance equation: $Q_{1} = W + 3W = 4W$.The efficiency $\eta$ of the cycle is defined as the ratio of work done to the heat absorbed: $\eta = \frac{W}{Q_{1}}$.Substituting the value of $Q_{1}$,we get $\eta = \frac{W}{4W} = \frac{1}{4} = 0.25$.

$A$ thermodynamic cycle takes in heat energy at a high temperature and rejects energy at a lower temperature. If the amount of energy rejected at the low temperature is $3$ times the amount of work done by the cycle,the efficiency of the cycle is

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