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11.Thermodynamics
normal

$30°C$ અને  $0°C $ ની વચ્ચે રેફ્‍જિરેટરનો પરફોર્મન્સ ગુણાંક

A

$10$

B

$1$

C

$9$

D

$0$

Solution

$\beta = \frac{{{T_2}}}{{{T_1} – {T_2}}} = \frac{{273^\circ C}}{{303^\circ C – 273^\circ C}} = \frac{{273^\circ C}}{{30^\circ C}} \approx 9$

Standard 11
Physics

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