In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from –3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>Temperature of the source is 27⁰C</p><p>T₁= 27+273= 300K</p><p>T₂= -3+273= 270K</p><p>Efficiency of heat engine = 1-T₂/T₁= 1-270/300=1/10</p><p>Efficiency of refrigerator&nbsp; is 50% of a perfect engine</p><p>= 0.5<span contenteditable="false"> <math> <mo>×</mo> <mi>η</mi> </math> </span>= 1/20</p><p>Coefficient of performance of the refrigerator<span contenteditable="false"> <math> <mi>β</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>Q</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>W</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mrow> <mrow> <mi>η</mi> </mrow> </mrow> <mrow> <mrow> <mi>'</mi> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>η</mi> </mrow> </mrow> </mfrac> </math> </span></p><p>=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <mn>1</mn> <mo>/</mo> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>20</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>19</mn> <mo>/</mo> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>20</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>19</mn> </math> </span></p><p>Q₂=<span contenteditable="false"> <math> <mi>β</mi> <mi>W</mi> </math> </span>=19W</p><p>= 19<span contenteditable="false"> <math> <mo>×</mo> </math> </span>1KW=19KW= 19kJ/s</p>

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