Q.During a cyclic process, a heat engine absorbs 600 J of heat from a hot reservoir, does work and ejects an amount of heat 200 J into the surroundings (cold reservoir). Calculate the efficiency of the heat engine.

A heat engine is any device that takes in heat, does some useful work, and then dumps the leftover heat somewhere else. Your car engine, a steam turbine, even your own body — all are heat engines.

The Intuition: You Can't Get Something for Nothing

Imagine you buy a bag of coal. You burn it to get heat. You want that heat to push a piston and turn a wheel. But you quickly notice: the engine gets hot, the exhaust pipe gets hot, the radiator gets hot. Some of that coal's heat is always wasted. You never turn 100% of the heat into motion.

Why? Because heat naturally flows from hot to cold. To make it do work, you have to let some of it "fall" to a lower temperature. The work you get is the difference between what went in and what came out. That difference is the efficiency.

The Precise Statement

Let’s label the quantities:

• $Q_h$ = heat absorbed from the hot reservoir (the fuel, the furnace, the sun)
• $Q_c$ = heat rejected to the cold reservoir (the exhaust, the cooling water, the atmosphere)
• $W$ = useful work output

By energy conservation (the First Law of Thermodynamics):

The efficiency $\eta$ is simply the fraction of the input heat that becomes work:

This is the definition. It is always less than 1 because $Q_c$ is always greater than zero — you must reject some heat.

What This Tells You

• To increase efficiency, you want $Q_c$ as small as possible — dump as little heat as possible. But you can never make $Q_c = 0$; that would violate the Second Law of Thermodynamics.
• The ratio $Q_c/Q_h$ is the fraction of input heat that is wasted. If you waste half, $\eta = 0.5$ (50%).
• Real engines (car engines, power plants) have efficiencies around 25–40%. The rest is lost as heat to the environment. …