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Standard XII

Chemistry

Heat of Reaction

In a Carnot's...

Question

In a Carnot's cycle, the working substance absorbs heat $Q_{1}$ at temperature $T_{1}$ and rejects heat $Q_{2}$ at temperature $T_{2}$ . The change of entropy during the Carnot's cycle is :

\frac{Q_{1}}{T_{1}}

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\frac{Q_{2}}{T_{2}}

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\frac{Q_{1}}{T_{1}} + \frac{Q_{2}}{T_{2}}

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Zero

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Solution

The correct option is D Zero
Change in entropy during heat absorption, $Δ S_{1} = Q_{1} / T_{1}$
Change in entropy during heat rejection, $Δ S_{2} = - Q_{2} / T_{2}$

For Carnot's cycle, $Q_{1} / T_{1} = Q_{2} / T_{2}$
Therefore net change in entropy is $Δ S_{1} + Δ S_{2} = 0$

Option D is correct.

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In a Carnot's cycle, the working substance absorbs heat Q1 at temperature T1 and rejects heat Q2 at temperature T2. The change of entropy during the Carnot's cycle is :

The correct option is D ZeroChange in entropy during heat absorption, ΔS1=Q1/T1Change in entropy during heat rejection, ΔS2=−Q2/T2For Carnot's cycle, Q1/T1=Q2/T2Therefore net change in entropy is ΔS1+ΔS2=0Option D is correct.

In a Carnot's cycle, the working substance absorbs heat $Q_{1}$ at temperature $T_{1}$ and rejects heat $Q_{2}$ at temperature $T_{2}$ . The change of entropy during the Carnot's cycle is :