COMEDK · Physics · 13. Thermodynamics
Carnot heat engine takes \(300 \mathrm{~J}\) of heat from a source at \(627^{\circ} \mathrm{C}\) and gives some part of it to \(\operatorname{sink}\) at \(27^{\circ} \mathrm{C}\). Work done by engine in one cycle is
- A \(200 \mathrm{~J}\)
- B \(300 \mathrm{~J}\)
- C \(150 \mathrm{~J}\)
- D \(120 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(200 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
The efficiency of a Carnot engine is given by \(\eta=1-\frac{T_{2}}{T_{1}}\) Given, \(\quad T_{1}=627^{\circ} \mathrm{C}=627+273=900 \mathrm{~K}\) and \(\quad T_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}\)…
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