Question

The efficiency of a Carnot engine at 7 °C (source temperature) is 50%. At what temperature (source temperature) the efficiency will be 70%? (Assume constant sink temperature)
1. 100 K
2. 366.7 K
3. 466.7 K
4. 466.7 °C

Answer 1

Correct Answer - Option 3 : 466.7 K

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.
  • 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.
  • 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 QRis heat rejected.

CALCULATION:

Given that:

T_H = 7 °C = 273 + 7 = 280 K

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

η = 1 - TC/TH = 50% = 0.5 = 1/2

1 - T_C/280 = 1/2

T_C = 280/2 = 140 K

Now η = 70% = 0.7

η = 1 - TC/TH

0.7 = 1 - 140/T'_H 

140/T'_H = 1 - 0.7 = 0.3

T_H = 140/0.3 = 466.7 K

• Hence option 3 is correct.