Path | \[\Delta Q\](Heat supplied) | \[\Delta U\](Increase in internal energy) | Work done by system on surrounding |
\[A\to B\] | 600 J | 200 J | 400 J |
\[B\to C\] | -100 J | 100 J | -200 J |
\[C\to A\] | -100 J | -300 J | 200 J |
A) 400 J, 100%
B) 600 J, 66.67%
C) 400 J, 66.67%
D) 600 J, 100%
Correct Answer: C
Solution :
Heat of cycle.= Net heat= 600-200 = 400 J Efficiency\[\eta =\frac{\text{Work}\,\text{done}}{\text{Heat}\,\text{supplied}}=\frac{400}{600}\]\[=66.67%\]You need to login to perform this action.
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