A Carnot's engine has an efficiency of 50 % at sink temperature `50^(@)C` .Calculate the temperature of source.

Options

A. `133^(@)C`
B. `143^(@)C`
C. `100^(@)C`
D. `373^(@)C`

Answer

D

Explanation

Thermal efficiency of a Carnot engine
`therefore" " eta=1-(T_(2))/(T_(1))` …..(i)
Wher e , `T_(2)` is temperature of sink and `T_(1)` is the temperature of the source.
Here, `eta=50%,T_(2)=50^(@)C=273+50=323K`
Putting these values in Eq .(i),We get
`(500)/(100)=1-(323)/(T_(1))`
`"or" " " (323)/(T_(1))=1-(1)/(2)=(1)/(2)`
`therefore" " T_(1)=646 K`
`=373^(@)C`