Home
Class 12
PHYSICS
A Carnot engine, having an efficiency of...

A Carnot engine, having an efficiency of `eta= 1/10` as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

A

99 J

B

100 J

C

1 J

D

90 J

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Similar Questions

Explore conceptually related problems

A carnot engine having an efficiency of (1)/(5) is being used as a refrigerator. If the work done on the refrigerator is 8 J, the amount of heat absorbed from the reservoir at lower temperature is:

A carnot engine having an efficiency of 1/4 is being used as a refrigerator. If the work done on the refrigerator is 5 J, the amount of heat absorbed from the reservoir at lower temperature is:

A carnot engine having an efficiency of 1/4 is being used as a refrigerator. If the work done on the refrigerator is 5 J, the amount of heat absorbed from the reservoir at lower temperature is:

A carnot engine having an efficiency of (1)/(5) is being used as a refrigerator. If the work done on the refrigerator is 8 J, the amount of heat absorbed from the reservoir at lower temperature is:

A Carnot heat engine has an efficiency of 10% . If the same engine is worked backward to obtain a refrigerator, then find its coefficient of performance.

In a carnot engine, when heat is absorbed from the source, its temperature

Assertion : Efficiency of a Carnot engine increase on reducing the temperature of sink. Reason : The efficiency of a Carnot engine is defined as ratio of net mechanical work done per cycle by the gas to the amount of heat energy absorbed per cycle from the source.

A system is provided 50 J of heat and work done on the system is 10 J. The change in internal energy during the process is

If the work done by a Carnot engine working between two temperatures 600K & 300K is used as the work input in Carnot refrigerator, working between 200K & 400K, find the heat (in J) removed from the lower temperature by refrigerator? The heat supplied to engine in 500J

The efficiency of a heat engine is defined as the ratio of the mechanical work done by the engine in one cycle to the heat absorbed from the high temperature source . eta = (W)/(Q_(1)) = (Q_(1) - Q_(2))/(Q_(1)) Cornot devised an ideal engine which is based on a reversible cycle of four operations in succession: isothermal expansion , adiabatic expansion. isothermal compression and adiabatic compression. For carnot cycle (Q_(1))/(T_(1)) = (Q_(2))/(T_(2)) . Thus eta = (Q_(1) - Q_(2))/(Q_(1)) = (T_(1) - T_(2))/(T_(1)) According to carnot theorem "No irreversible engine can have efficiency greater than carnot reversible engine working between same hot and cold reservoirs". Efficiency of a carnot's cycle change from (1)/(6) to (1)/(3) when source temperature is raised by 100K . The temperature of the sink is-