Question

A heat engine operating between the

temperatures T1 and T2 takes in Qı calories at
temperature T1 and gives up Q2 calories at
temperature T2. The efficiency of this heat
engine is
(Q1 - Q2)/Q2.
(Q1 - Q2)/Q1.
(T2 - T1/T2.
Q2/(Q1 - Q2).
O T1/(T1-T2).

Answer 1

(Q1 - Q2)/Q1

Step-by-step explanation:

The efficiency of any device can be given by the following formula:

`Efficiency = \eta` = Output/Input

Now, for engine the output is the mechanical work that it does and the input is the heat that it absorbs. Let:

W = Work Done by the engine

Q₁ = Heat Absorbed by the System

Q₂ = Heat Rejected by the System (Because According to 2nd Law of Thermodynamics, engine can not convert the entire heat into work without rejection of some heat)

Therefore, the equation becomes:

`\eta = (W)/(Q_(1))`

but, the work can be written as the difference between the absorbed heat and the rejected heat. Hence,

`\eta = (Q_(1) - Q_(2))/(Q_(1))`

Therefore, the correct option is:

(Q1 - Q2)/Q1