Thermal Physics> a carnot engine whose sink is at 90 k has...

a carnot engine whose sink is at 90 k has an efficiency of 30%. by how much the teperature of the source be increased to have its efficiency equal to 50%, keeping sink teperature constant

raja , 7 Years ago

Grade 11

2 Answers

Khimraj

efficiency

$\eta$ = 1 – T_sink/T_source

$\eta$ +1 = T_sink/T_source

As T_source is constant

( $\eta$ ₂ +1)/( $\eta$ ₁ +1) = T_sink2/T_sink1

1.6/1.3 = x/90

x = 110.76

%increase in teperature of the source = 20.76*100/90 = 23%

Hope it clears.....................................

Last Activity: 7 Years ago

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Thermal Physics> a carnot engine whose sink is at 90 k has...

a carnot engine whose sink is at 90 k has an efficiency of 30%. by how much the teperature of the source be increased to have its efficiency equal to 50%, keeping sink teperature constant

raja , 7 Years ago

Khimraj

Aastha Singh

LIVE ONLINE CLASSES

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