Question

calcutale thermal efficency of heat engine that works between 227⁰C and 27⁰c. the thermal efficiency of heat engine can never be 100%. why?

Answer 1

Required efficiency is, η = 1 – (temperature of sink)/(temperature of source)

=> η = 1 – (27 + 273)/(227 + 273) = 0.4

And in percentage it is, η = 0.4 × 100 = 40%

Explanation of efficiency of heat engine or thermal engine or carnot’s engine:

In a Carnot’s engine an ideal gas enclosed within a cylinder fitted with a piston which is is first isothermally expanded by letting some heat flow from the source, then it is adiabatically compressed where no heat is exchanged with the surrounding. After adiabatic compression the system is kept in the sink and is allowed to expand isothermally where it gives off a part of heat to the sink. Then the system is again compressed adiabatically. This adiabatic compression brings the system back to the initial state.

Let Q₁ amount of heat be taken from the source at temperature T₁

Let Q₂ amount of heat be expelled from the system to the sink kept at temperature T_2.

Thus work done by the engine must be equal to W = Q₁ – Q₂

Now the efficiency of the engine η = Work done /Input heat

Thus η = (Q₁ – Q₂)/Q₁ = 1 – Q₂/Q₁

Now Q_2­ = ST₂ and Q₁ = ST₁ Where S = entropy of the engine (assumed to be constant)

=> η = 1 – T₂/T₁ = (T₁ – T₂ )/T₁

Since, the numerator here is always less than the denominator, η is always less than 1. Hence, the efficiency is never 100%.