a carnot engine efficiency of 50 % when it sinks temperature 27°C the temperature of the source is
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Efficiency = 50%
Description of efficiency as a fraction would give
Efficiency = 1/2
Now,
For a Carnot Engine
Efficiency = 1-Temperature of Sink/Temperature of source
(You may find the derivation in BM Sharma(Physics for JEE Advanced or Any where on web)
Temperature must be kept in Kelvin units
So,
1/2 = 1- (273+27)/Temperature of source
So,
Temperature of source = 600 Kelvin = 327° C
Description of efficiency as a fraction would give
Efficiency = 1/2
Now,
For a Carnot Engine
Efficiency = 1-Temperature of Sink/Temperature of source
(You may find the derivation in BM Sharma(Physics for JEE Advanced or Any where on web)
Temperature must be kept in Kelvin units
So,
1/2 = 1- (273+27)/Temperature of source
So,
Temperature of source = 600 Kelvin = 327° C
sneha14640:
a)300 b) 327 c)373 d)273
Options are in degree celsius, So, finally answer will be (600-273)° C = 327°Celsius
thanks
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Answered by
4
Temperature of source is 327°C
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