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11. thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics11. thermodynamics1 Mark
MCQ
A  refrigerator works between $4^o C$ and $30^o C.$  It is required to remove $600$ calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is ....... $W$ (Take $1\, cal \,=\, 4.2\, Joules\,)$ 
  • A
    $23.65$
  • ✓
    $236.5$
  • C
    $2365$
  • D
    $2.365$
✓

Answer

Correct option: B.
$236.5$
b
$Given,\,{T_2} = {4^ \circ }C = 277\,K,{T_1} = {30^ \circ }C = 303\,K$

${Q_2} = 600\,cal\,per\,second$

Coefficient of performance, $\alpha  = \frac{{{T_2}}}{{{T_1} - {T_2}}}$

$ = \frac{{277}}{{303 - 277}} = \frac{{277}}{{26}}$

Also,$\alpha  = \frac{{{Q_2}}}{W}$

$\therefore $ $Work\,to\,be\,done\,per\,second=power\,required$

$ = W = \frac{{{Q_2}}}{\alpha } = \frac{{26}}{{277}} \times 600\,cal\,per\,second$

$ = \frac{{26}}{{277}} \times 600 \times 4.2\,J\,per\,second = 236.5\,W$

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