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JEE Main 2021 (25 Jul, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2021 (25 Jul, Shift 2)

Q45JEE Main 2021 · 25 JUL · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

A heat engine has an efficiency of 16\frac{1}{6}. When the temperature of sink is reduced by 62∘^\circ C, its efficiency gets doubled. The temperature of the source is:

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Answer: (d)

  1. T2=56T1T_2 = \frac{5}{6}T_1 and, with doubled efficiency 13\frac{1}{3}, T2−62=23T1T_2 - 62 = \frac{2}{3}T_1.
  2. Subtract: T16=62⇒T1=372 K\frac{T_1}{6} = 62 \Rightarrow T_1 = 372\text{ K}.
  3. T1=372−273=99∘CT_1 = 372 - 273 = 99^\circ\text{C}
  4. So the answer is (d) 99∘C99^\circ\text{C}.
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