Gujarati
Hindi
11.Thermodynamics
normal

An ideal gas heat engine operates in Carnot's cycle between $227\,^oC$ and $127\,^oC$ . It absorbs $6.0 \times 10^4\,cal$ at higher temperature. The amount of heat converted into work is equal to

A

$4.8 \times 10^4\,cal$

B

$3.5 \times 10^4\,cal$

C

$1.6 \times 10^4\,cal$

D

$1.2 \times 10^4\,cal$

Solution

$\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}}$

$1-\frac{400}{500}=\frac{W}{6 \times 10^{4}}$

$\mathrm{W}=1.2 \times 10^{4} \mathrm{cal}$

Standard 11
Physics

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