A Carnot engine , having an efficiency of n= 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
Answers
Efficiency of Carnot's Engine can be written after deriving it as:
n = 1-Q2/Q1
Where,
n is the efficiency,
Q1 is the absorbed or heat gained,
Q2 is the unused or rejected heat.
Now according to the given conditions,
Efficiency is 10% or 0.1,
And the engine is used as a refrigerator due to which it must absorb the heat from the coldr body(specifically called sink in thermodynamics) and transfer t to the hot body (called source) to lower its temperature ( which is the principle of refrigerator).
Hence the work that is done lies somewhere between the gained and the rejected heat as it is clear that work is done by the expense of the heat that is gained. So the rejected heat is of no use, so called unused heat.
Hence we can say that the work is the difference of the gained and rejected heat.
W= Q1-Q2
Which is:
10 = Q1-Q2
Q2 = Q1-10
Now citing the formula:
n = 1-Q2/Q1
Taking lcm:
n = Q1-Q2/Q1
Put the values:
0.1 = Q1-(Q1-10)/Q1-10
0.1= Q1-Q1+10/Q1-10
Q1-10= 100
Q1=110J