Question

A reversible heat engine absorbs 20 kJ of heat from a source at 500 K and dissipates it to the reservoir at 400 K. The efficiency of the heat engine is _______%. 

Answer 1

Concept:

• Carnot engine: The theoretical engine which works on the Carnot cycle is called a Carnot engine.
  • It gives the maximum possible efficiency among all types of heat engines.

• Heat source: The part of the Carnot engine which provides heat to the engine is called a heat source.
  • The temperature of the source is maximum among all the parts.

• Heat sink: The part of the Carnot engine in which an extra amount of heat is rejected by the engine is called a heat sink.

• The amount of work that is done by the engine is called work done.

The efficiency (η)of a Carnot engine is given by:

\(η = 1 - \frac{{{T_C}}}{{{T_H}}} = \;\frac{{Work\;done\left( W \right)}}{{{Q_{in}}}} = \;\frac{{{Q_{in}} - \;{Q_R}}}{{{Q_{in}}}}\)

Where TC is the temperature of the sink, TH is the temperature of the source, W is work done by the engine, Qin is the heat given to the engine/heat input and QR is heat rejected.

Calculation:

Given:
The temperature of the source T_H = 500K

The temperature of the sink T_c = 400K

The efficiency is given by:

\(η = 1 - \frac{T_{2}}{T_{1}}\)

Substituting the values in the above equation, we get:
\(η = 1 - \frac{400}{500}\)

or, η = .2, converting it in percentage, we get: .2 × 100 = 20%

Hence the efficiency is 20%.