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Heat and Thermodynamics question

2018 · Q138
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Heat and Thermodynamics question

2018 · Q138

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
  1. A
    26.8%
  2. B
    20%
  3. C
    6.25%
  4. D
    12.5%
View written solutionFree

Correct answer: A

Efficiency of ideal heat engine,

η=(1−T2T1)\eta = \left( {1 - {{{T_2}} \over {{T_1}}}} \right)η=(1−T1​T2​​)

Freezing point of water = 0oC = 273 K

Boiling point of water = 100oC = (100 + 273) K = 373 K

∴\therefore∴ Sink temperature, T2 = 0oC = 0 + 273 = 273 K

∴\therefore∴ Source temperature, T1 = 100oC = 100 + 273 = 373 K

Percentage efficiency, %η=(1−T2T1)×100\eta = \left( {1 - {{{T_2}} \over {{T_1}}}} \right) \times 100η=(1−T1​T2​​)×100

= (1−273373)×100\left( {1 - {{273} \over {373}}} \right) \times 100(1−373273​)×100

= (100373)×100\left( {{{100} \over {373}}} \right) \times 100(373100​)×100 = 26.8%

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