A Carnot engine works between the temperature 227° C and 127° C. If the work output of the engine is 10⁴ J, then the amount of heat rejected to the sink will be:

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Official Sr. Teacher Gr II NON-TSP Science (Held on : 1 Nov 2018)

Answer 1

CONCEPT:

Carnot engine an ideal reversible engine that operates between two temperatures T1 (Source), and T2 (Sink).
Carnot engine operates through a series of two isothermal and adiabatic processes called the Carnot cycle.
The steps of the Carnot cycle are
1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression
The efficiency of the Carnot engine is defined as the ratio of net work done per cycle by the engine to heat absorbed per cycle by the working substance from the source.
The efficiency is given by

\(\eta = \frac{W}{Q_1} =\frac{Q_1-Q_2}{Q_1} = 1-\frac{Q_2}{Q_1}\)

Where W = Work, Q1 = Amount of heat absorbed, Q2 = Amount of heat rejected

As \(\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\)

\( \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

Where T1 = temperature of the source and T2 = temperature of the sink.

SOLUTION:

Given - T₁ = 227° C = 500 K, T₂ = 127° C = 400 K and W = 104 J

\(⇒ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

\(⇒ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{{{400}}}{{{500}}}=\frac{1}{5}\)

\(⇒ Q_1= \frac{W}{\eta}=\frac{10^4}{\frac{1}{5}}=5×10^4\, J\)

⇒ Q₂ = Q₁ - W

⇒ Q2 = 5 × 10⁴ - 1 × 10⁴ = 4 × 10⁴ J