STATE AND THE LAWS OF THERMAL CONDICTIVITY DESCRIBE THE WORKING OF CARNOT ENGINE.

The rate of flow of heat, i.e. ΔQ/Δt, through a metal rod of area of cross-section ‘A’, length ‘L’ temperature difference ΔT is given by,

ΔQ/Δt = kAΔT/L

Where, ‘k’ is the thermal conductivity of the material of the rod.

Carnot’s engine is a heat engine that works between two temperatures T₁ and T₂ respectively of source and sink which is the most efficient heat engine possible. Source is a place with high temperature (T₁) and the sink is a place having low temperature (T₂). By letting heat flow from source to sink through the engine a part of work is converted into work by the 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%.