Question

A carnot engine having an efficiency of 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 :​

Answer 1

Solution :

Given,

Efficiency of the engine = 1/10

Workdone = 10J

By using the formula of the efficiency of engine

Efficiency = W /q

Now,

Efficiency = W /q

=> W/q1 = 1/10

=> 10/q1 = 1/10

=> q1 = 100 J

.°. q1 = 100J

Now,

=> W = q1 - q2

=> W + q2 = q1

=> 10 + q2 = 100

=> q2 = 100 - 10

=> q2 = 90J

.°. q2 = 90J

Thus, the amount of energy absorbed from the reservoir at lower temperature is 90J.

Answer 2

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