Question

The efficiency of a Carnot engine operating with reservoir temperatures of 100°C and -23°C will be
1. \(\frac {100 - 23}{100}\)
2. \(\frac {100 + 23}{100}\)
3. \(\frac {100 + 23}{373}\)
4. \(\frac {100 - 23}{373}\)

Answer 1

Correct Answer - Option 3 : \(\frac {100 + 23}{373}\)

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 which is done by the engine is called as work done.
• The efficiency (η)of a Carnot engine is given by:

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

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 QRis heat rejected.

CALCULATION:

T₁ = 100°C = 273 + 100 = 373 K

T₂ = -23°C = 273 - 23 = 250 K

• The efficiency of the Carnot engine: 

\(\Rightarrow \eta =\frac{W}{{{Q}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}=\frac{{{373}}-{{250}}}{{{373}}}=\frac{{{123}}}{{{373}}}=\frac{{{100}}+{{23}}}{{{373}}}\)

• Therefore the Efficiency of the Carnot cycle is option 3.