A carnot engine rejects 40% of the heat absorbed from a source to sink at 27°C. What is the value of source temperature in °C?

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A carnot engine rejects 40% of the heat absorbed from a source to sink at 27°C. What is the value of source temperature in °C?

1. 327
2. 54
3. 477
4. 600

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Correct Answer - Option 3 : 477

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:

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

Where Tis the temperature of the sink, Tis 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.

The relation between the Kelvin scale (K) and the celsius scale (°C);

K = °C + 273

CALCULATION:

Given:

Heat sink temperature (TC) = 27°C = 273 + 27 = 300 K, Qr = 0.4Qin

Heat source temperature (TH) = ?

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

$η = 1 - \frac{{{300}}}{{{T_H}}} = \;\frac{{{Q_{in}} - \;{0.4Q_{in}}}}{{{Q_{in}}}}=0.6$

$\frac{300}{T_H}=0.4, T_H= \frac{300}{0.4}=750~K$

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