1.

An ideal heat engine works between source at 127^(@)C and sink 27^(@)C. If 800 J heat is taken from reservoir, the amount of heat rejected to sink is

Answer»

300 J
400 J
500 J
600 J

Solution :`eta=1-(T_(2))/(T_(1))`
`eta=1-(Q_(2))/(Q_(1))=1-(T_(2))/(T_(1))=(300)/(600)=(1)/(2)`
`therefore (Q_(2))/(Q_(1))=(T_(2))/(T_(1))`
`Q_(2)=(T_(2))/(T_(1))xxQ_(1)=(300)/(400)xx800=600J`


Discussion

No Comment Found

Related InterviewSolutions