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11.Thermodynamics
normal
An ideal 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
$ = \frac{{500 – 400}}{{500}} = \frac{1}{5} = \frac{W}{{6 \times {0^{ – 4}}}}\,\,\,W = 1.2 \times {10^4}\,cal$
Standard 11
Physics
