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A car not's engine whose sink is at a temperature of 300K has an efficiency of 40% By space should the temperature of the source be increase the efficiency to 60%
$n = 1 - { T_2 \over T_1 } $
An ideal gas heat engine is operating between $ 227 ^ \circ C and 127 ^\circ $ . It absorbs $10^4 J $ Of heat at the higher temperature. The amount of heat Converted into. work is................. J.
$n = 1 - { T_2 \over T_1 } = { 1- 400 \over 500 } = {1 \over 5} $ $ W = nQ_1 $
Efficiency of a car not engine is 50%, when temperature of outlet is 500K. in order to increase efficiency up to 60% keeping temperature of intake the same what is temperature of out let.
$ n = 1 - { T_2 \over T_1 }$, ${ T_2 \over T_1} should be minimum
A car not engine takes $ 3 \times 10^6 cal $ of heat from a reservoir at $ 627 ^\circ C $ and gives to a sink at $ 27 ^\circ C $. The work done by the engine is
$ (i) n = 1 - { T_2 \over T_1 }$ $ (ii) n^1 = 1 - { 2T_2 \over 2T_1 } = 1 - { T_2 \over T_1 } = n $
For which combination of working temperatures the efficiency of Car not's engine is highest.
The efficiency of a Carnot engine is given by the formula: $$\eta = 1 - \frac{T_c}{T_h}$$ where $T_c$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir. To maximize efficiency, the difference between $T_h$ and $T_c$ should be as large as possible. For the given options:
An ideal heat engine working between temperature $T_1 and T_2 $ has an efficiency n. The new efficiency if both the source and sink temperature are doubled, will be
The efficiency of an ideal heat engine is given by $$\eta = 1 - \frac{T_2}{T_1}$$ If both the source and sink temperatures are doubled, the efficiency becomes: $$\eta' = 1 - \frac{2T_2}{2T_1} = 1 - \frac{T_2}{T_1} = \eta$$ Therefore, the new efficiency remains the same as the original efficiency.
An ideal refrigerator has a freetes at a temperature of $- 13 ^\circ C $ . , The coefficent of performance of the engine is 5. The temperature of the air to which heat is rejected will be.
$ \alpha = { T_2 \over T_1 - T_2 } $
An engine is supposed to operate between two reservoirs at temperature $ 727 ^\circ C and 227 ^\circ C $ . The maximum possible efficiency of such an engine is
$n = 1 - { T_2 \over T_1 } = 1 - { 500 \over 1000} = { 1 \over 2}$
A car not engine Convertsm one sixth of the heat input into work. When the temperature of the sink is reduces by $ 62 ^\circ C $ the efficiency of the engine is doubled. The temperature of the source and sink are
$ (i) n = 1 - { T_2 \over T_1 } = { W \over Q_1} = { 1 \over 6} $ $ \therefore n = { 1 \over 6} - (1) $ $ (ii) n^1 = 1- { T_2 -62 \over T_1}$ $ = 1 - { T_2 \over T_1 } + { 62 \over T_1 } $ $ = n+ { 62 \over T_1 } - (2) $ $ Now , n^1 = 2n$
What is the value of sink temperature when efficiency of engine is 100%
$ n = 1 - { T_2 \over T_1 } $
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