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

Q: 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

d $\Delta H = mL = 5 \times 336 \times {10^3} = {Q_{sink}}$ $\frac{{{Q_{sink}}}}{{{Q_{source}}}} - \frac{{{T_{sink}}}}{{{T_{source}}}}$ $\therefore {Q_{source}} = \frac{{{T_{source}}}}{{{T_{sink}}}} \times {Q_{sink}}$ Energy consumed by freezer $\therefore {W_{output}} = {Q_{source}} - {Q_{sink}}$ $ = {Q_{sink}}\left( {\frac{{{T_{source}}}}{{{T_{sink}}}} - 1} \right)$ $Given:\,{T_{source}} = {27^ \circ }C + 273 = 300K,$ ${T_{sink}} = {0^ \circ }C + 273 = 273\,K$ ${W_{output}} = 5 \times 336 \times {10^3}\left( {\frac{{300}}{{273}} - 1} \right) = 1.67 \times {10^5}J$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$1.51 \times {10^5}\,J$
B$1.68 \times {10^6}\,J$
C$1.71 \times {10^7}\,J$
D$1.67 \times {10^5}\,J$

JEE MAIN 2016, Diffcult

STD 11 Science
NEET
STD 11 - 11. thermodynamics
Physics

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