Two Carnot engines A and B operate in series such that engine A absorbs heat at T? and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at T?. When work done in both the cases is equal, the value of T is:

Option 1 - <p>(2/3)T₁ + (1/3)T₃<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>(2/3)T₁ + (3/2)T₃<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>(1/2)T₁ + (1/3)T₃<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>(1/3)T₁ + (2/3)T₃</p>
3 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
V
6 months ago
Correct Option - 4
Detailed Solution:

η_A = W_A/Q? = 1 - T/T?
η_B = W_B/Q? ' = 1 - T? /T
W_A = Q? (1 - T/T? ) = Q? - Q?
W_B = Q? ' (1 - T? /T) = Q? /2 (1 - T? /T)
Given W_A = W_B
Q? (1 - T/T? ) = (Q? /2) (1 - T? /T)
Q? (T? /T) (1-T/T? ) = (Q? /2) (1-T? /T)
(T? /T - 1) = (1/2) (1-T? /T)
2T? /T - 2 = 1 - T? /T
2T? /T + T? /T = 3
T = (2T? +T? )/3

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Physics Thermodynamics 2025

Physics Thermodynamics 2025

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