A Carnot freezer takes heat from water at $0\,^oC$ inside it and rejects it to the room at a temperature of $27\,^oC$. The latent heat of ice is $336 \times 10^3\, J\,kg^{-1}$. lf $5\, kg$ of water at $0\,^oC$ is converted into ice at $0\,^oC$ by the freezer, then the energy consumed by the freezer is close to
JEE MAIN 2016, Diffcult
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$\Delta H = mL = 5 \times 336 \times {10^3} = {Q_{sink}}$
$\frac{{{Q_{sink}}}}{{{Q_{source}}}} - \frac{{{T_{sink}}}}{{{T_{source}}}}$
$\therefore {Q_{source}} = \frac{{{T_{source}}}}{{{T_{sink}}}} \times {Q_{sink}}$
Energy consumed by freezer
$\therefore {W_{output}} = {Q_{source}} - {Q_{sink}}$
$ = {Q_{sink}}\left( {\frac{{{T_{source}}}}{{{T_{sink}}}} - 1} \right)$
$Given:\,{T_{source}} = {27^ \circ }C + 273 = 300K,$
${T_{sink}} = {0^ \circ }C + 273 = 273\,K$
${W_{output}} = 5 \times 336 \times {10^3}\left( {\frac{{300}}{{273}} - 1} \right) = 1.67 \times {10^5}J$
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