Home
Class 11
PHYSICS
The efficiency of heat engines and refri...

The efficiency of heat engines and refrigerators, the direction of physical and chemical process, etc., are also studied in …

A

Nature Science

B

Cryogenics

C

Galaxies

D

Thermodynamics

Text Solution

Verified by Experts

The correct Answer is:
D
Thermodynamics
Promotional Banner

Topper's Solved these Questions

  • PHYSICAL WORLD

    KUMAR PRAKASHAN|Exercise SECTION-E (QUESTIONS FROM MODULE)|9 Videos

  • PHYSICAL WORLD

    KUMAR PRAKASHAN|Exercise Section - C ( Objective Questions)|40 Videos

  • OSCILLATIONS

    KUMAR PRAKASHAN|Exercise QUESTION PAPER (SECTION - D)|1 Videos

  • QUESTION ASKED IN NEET -2019

    KUMAR PRAKASHAN|Exercise MULTIPLE CHOICE QUESTIONS|9 Videos

Similar Questions

Explore conceptually related problems

What is the efficiency of heat engine ? Write it formula.

What is theoretical efficiency of heat engine ?

Why can't the efficiency of a heat engine ever be 100 % ?

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". An inventor claims to have developed an engine working between 600K and 300K capable of having an efficiency of 52% , then -

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-

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". A carnot engine whose low temperature reservoir is at 7^(@)C has an efficiency of 50% . It is desired to increase the efficiency to 70% . By how many degrees should the temperature of the high temperature reservoir be increased?

The efficiency of an ideal heat engine working between the freezing point and boiling point of water is ____

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 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 10J, the amount of energy absorbed from the reservoir at lower temperature is

What would be the efficiency of the heat engine diagramed as shown below?