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An ideal gas heat engine operates in Carnot cycle between 227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C} . It absorbs 6 \times 10^{4} \mathrm{cal} of heat at higher temperature. Amount of heat converted to work is

Option: 1

2.4 \times 10^{4} \mathrm{cal}


Option: 2

6 \times 10^{4} \mathrm{cal}


Option: 3

1.2 \times 10^{4} \mathrm{cal}


Option: 4

4.8 \times 10^{4} \mathrm{cal}


Answers (1)

best_answer

From the relation,
\mathrm{\frac{Q_{2}}{Q_{1}}=\frac{T_{2}}{T_{1}}}

\mathrm{Given, Q_{1}=6 \times 10^{4} \mathrm{cal}}

\mathrm{T_{1}=227+273 =500 \mathrm{~K}}
\mathrm{T_{2}=127+273 =400 \mathrm{~K} }

\mathrm{\therefore \quad \frac{Q_{2}}{2 \times 10^{4}} =\frac{400}{500} }

\mathrm{\Rightarrow \quad Q_{2} =\frac{4}{5} \times 6 \times 10^{4}=4.8 \times 10^{4} \mathrm{cal} }

Now, t converted to work
\mathrm{=Q_{1}-Q_{2} =6.0 \times 10^{4}-4.8 \times 10^{4} }
                        \mathrm{=1.2 \times 10^{4} \mathrm{cal}}

Posted by

Ajit Kumar Dubey

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