A heat engine is involved with exchange of heat of 1915 J, –40 J, +125 J and –QJ, during one cycle achieving and efficiency of 50.0%. The value of Q is:

Option 1 - <p>40 J<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>640 J</p>
Option 3 - <p>400 J</p>
Option 4 - <p>980 J</p>
6 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
6 months ago
Correct Option - 4
Detailed Solution:

Efficiency η = Q_net/Q? = (1915-40+125-Q)/ (1915+125) = 0.5.
(2000-Q)/2040=0.5 ⇒ Q=980J.

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Physics Thermodynamics 2025

Physics Thermodynamics 2025

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