A heat engine absorbs heat Q_1 at temperature | vedclass

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(D) The $1^{st}$ law of thermodynamics states that the change in internal energy $\Delta U = q + w$. For a cyclic process,$\Delta U = 0$,so $q = -w$.

Vedclass · Accepted Answer

Does not violate $1^{st}$ law of thermodynamics

Vedclass · Answer

(D) The $1^{st}$ law of thermodynamics states that the change in internal energy $\Delta U = q + w$. For a cyclic process,$\Delta U = 0$,so $q = -w$. Here,the total heat absorbed is $Q_1 + Q_2$ and the work done is $W = Q_1 + Q_2$. Since $W = Q_{total}$,the $1^{st}$ law is satisfied. However,this process violates the $2^{nd}$ law of thermodynamics,which states that heat cannot be completely converted into work in a cyclic process without any other effect.

$A$ heat engine absorbs heat $Q_1$ at temperature $T_1$ and heat $Q_2$ at temperature $T_2$. Work done by the engine is $(Q_1 + Q_2)$. This data

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