An engine absorbs heat Q_1 at temperature T_1 | vedclass

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(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta U = q + w$. For a cyclic process,$\Delta

Vedclass · Accepted Answer

The first law of thermodynamics is not violated.

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(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta U = q + w$. For a cyclic process,$\Delta U = 0$,which implies $q = -w$ or $w = -q$. Here,the total heat absorbed is $q_{total} = Q_1 + Q_2$. Therefore,the work done by the engine should be $W = -(Q_1 + Q_2)$. The given expression $W = Q_1 + Q_2$ implies $W = -q_{total}$,which is consistent with the first law of thermodynamics for a cyclic process where the system returns to its initial state. Thus,the first law is not violated.

An engine absorbs heat $Q_1$ at temperature $T_1$ and heat $Q_2$ at temperature $T_2$. The work done by the engine is given as $W = Q_1 + Q_2$. Which of the following statements is correct?

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