**Heat Engines**

A device which transforms heat energy partially into work or mechanical energy is called as heat engine.

The heat engine operates in a cyclic process. During this process, the engine takes in energy by heat from a high-temperature energy reservoirs, does work and then takes out energy by heat to a lower-temperature reservoir.

Below is a schematic diagram of how a heat engine works.

**Figure**

First, the engine acquires heat denoted as \(|Q_h|\) from a hot reservoir. Then, it does work denoted as \(W_{eng}\) and after, gives off an amount of energy \(|Q_c|\) to the cold reservoir.

Since the engine is working in a cyclic process, the initial and final internal energies are equal, which makes the total internal energy zero, that is

\(\Delta E_{int} = 0\).

Note that from the first law of thermodynamics,

\(\Delta E_{int} = Q + W = Q - W_{eng} = 0\)

and the work, \(W_{eng}\) done by the engine is equal to the total heat, \(Q_{net}\) transferred to it,

\(W_{eng}=Q_{net}\).

Based on the figure above, the total heat transferred to the engine is equal to the difference between the absorbed heat, \(|Q_h|\) and expelled heat, \(|Q_c|\) which is expressed mathematically as

\(Q_{net} = |Q_h| - |Q_c|\).

From these relationships, the equation for the work done by the can be summarized as

\(W_{eng} = |Q_h| - |Q_c|\) (1)

The thermal efficiency of the heat engine is the ratio of the work done by the engine in one cycle and the its absorbed energy at the higher temperature of the cycle. It can be written mathematically as

\(e = {W_{eng} \over |Q_h|}\) (2)

Replacing equation (1) to \(W_{eng}\) in equation (2), we have

\(e = {{|Q_h| - |Q_c|} \over |Q_h|}\)

which can be simplified as

\(e = 1 - {|Q_c| \over |Q_h|}\) (3)

Practically, all heat engines gives off only partial amount of the absorbed energy \(Q_h\) by mechanical work, and thus, they have efficiency less than 100%.

The absolute values are used to make all transfer of energy by heat positive. The direction of the transfer will be specified by a positive or negative sign.

**Example.**

A heat engine transfers an energy of \(3.5 \times 10^2\; J\) from a hot reservoirs during a cycle and gives off \(2.2 \times 10^2\; J\) of energy to the cold reservoir.

- How much work is done by the engine?
- What is the efficiency of the engine?

Solution:

- The work done by the engine is the difference between the absorbed energy and the expelled energy. To solve, we use equation (1)

\(W_{eng} = |Q_h| - |Q_c|\)

Substitute the given values,

\(W_{eng} = 3.5 \times 10^2\; J - 2.2 \times 10^2\; J = 1.3 \times 10^2\; J\)

- To solve for the efficiency of the engine, we can use either of equation (2) or equation (3).

b.1. Using equation (2),

\(e = {W_{eng} \over |Q_h|} = {1.3 \times 10^2\; J \over 3.5 \times 10^2\; J} = 0.37\).

b.2. Using equation (3),

\(e = 1 - {|Q_c| \over |Q_h|} = 1 - {2.2 \times 10^2\; J \over 3.5 \times 10^2\; J } = 0.37\).