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 :
[c] Heat of cycle = Net heat |
\[=600-200=400\text{ }J\] |
Efficiency \[\eta =\frac{\text{Work}\,\text{done}}{\text{Heat}\,\text{supplied}}=\frac{400}{600}\] |
\[=66.67%\] |
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