JEE Mains · Physics · STD 11 - 11. thermodynamics
A heat engine operates between a cold reservoir at temperature \({T}_{2}=400\, {K}\) and a hot reservoir at temperature \({T}_{1} .\) It takes \(300 \,{J}\) of heat from the hot reservoir and delivers \(240\, {J}\) of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be \(....{K}\)
- A \(400\)
- B \(500\)
- C \(300\)
- D \(100\)
Answer & Solution
Correct Answer
(B) \(500\)
Step-by-step Solution
Detailed explanation
\({Q}_{\text {in }}=300\, {J} ; Q_{\text { out }}=240\, {J}\) Work done \(=Q_{\text {in }}-Q_{\text {out }}=300-240=60 {J}\) Efficiency \(=\frac{W}{{Q}_{\text {in }}}=\frac{60}{300}=\frac{1}{5}\) efficiency \(=1-\frac{{T}_{2}}{{T}_{1}}\)…
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