An engine is working . It takes 100 calories of heat from source and leaves 80 calories of heat to sink . If the temperature of source is `127^(@)` C, then temperature of sink is

Options

A. `147^(@)C`
B. `47^(@)C`
C. `100^(@)C`
D. `47K`

Answer

B

Explanation

Heat taken , from source `Q_(1)=100` cal
Heat left to sink `Q_(2)=80` cal
`therefore ` efficiency of the engine `=eta=1-(Q_(2))/(Q_(1))=1-(80)/(100)`
Temperature of the source `T_(1)=127^(@)C=400 K`
Temperature of the sink `T_(2)= ?`
We know that
`eta=1-(T_(2))/(T_(1))`
`implies 0.2 =1-(T_(2))/(400)`
`implies (T_(2))/(400)=1-0.2=0.8`
`implies T_(2)=320 K=47^(@)C`