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11.Thermodynamics
normal
An ieal heat engine operates on Carnot cycle between $227\,^oC$ and $127\,^oC$. It absorbs $6 \times 10^4\, cal$ at the higher temperature. The amount of heat converted into work equals 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
$\mathrm{n}=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}$
$\Rightarrow \frac{100}{500}=\frac{\mathrm{W}}{6 \times 10^{4}} \Rightarrow \mathrm{W}=1.2 \times 10^{4} \mathrm{cal}$
Standard 11
Physics
