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 $(Q_{1} + Q_{2}) J$ . This data:

violates

1^{s t}

law of thermodynamics

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violates

1^{s t}

law of thermodynamics

Q_{1}

-

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violates

1^{s t}

law of thermodynamics

Q_{2}

-

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does not violates

1^{s t}

law of thermodynamics

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Solution

The correct option is D does not violates $1^{s t}$ law of thermodynamics
According, First Law of Thermodynamics,

Change in $E = Q +$ work done

Change in internal energy is zero for heat engines because initial and final states are same.

Total work done $= -$ Total heat exchange $= - (Q_{1} + Q_{2})$ .

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In one complete cycle, a heat engine extracts heat $Q_{h}$ from a thermal reservoir, does work W and does not eject any heat into the environment. This heat engine is impossible because its operation violates:

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and rejects heat

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. The change of entropy during the Carnot's cycle is :

Q. According to 1st law of thermodynamics :

Q. 1st law of Thermodynamics can explain

According to 1st law of Thermodynamics

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A heat engine absorbs heat Q1 at temperature T1 and heat Q2 at temperature T2 . Work done by the engine is (Q1+Q2)J . This data:

The correct option is D does not violates 1st law of thermodynamicsAccording, First Law of Thermodynamics,Change in E=Q+ work doneChange in internal energy is zero for heat engines because initial and final states are same.Total work done =−Total heat exchange =−(Q1+Q2).

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 $(Q_{1} + Q_{2}) J$ . This data: