question_answer 30)

                  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 1 kW power, and heat is transferred from \[-{{3}^{o}}C\] to \[{{27}^{o}}C\], find die beat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.                

Answer:

                  Here, \[{{T}_{2}}=273-3=270\text{ }K\]                 \[{{T}_{1}}=273+27=300\text{ }K\]                 \[P=\frac{W}{t}=1\,kW\,=\,1000\,J{{s}^{-1}}\,=1\,kJ\,\,{{s}^{-1}}\]                 Efficiency of perfect engine, \[\eta \,=1-\frac{{{T}_{2}}}{{{T}_{2}}}\]                 \[=1-\,\frac{270}{300}\,=0.1\]                 Efficiency of refrigerator, \[{{\eta }_{x}}=\,50%\,\eta \]                 \[=\,\frac{1}{2}\times 0.1\,\,=0.05\]                 Now, let Q be the heat transferred at higher temperature per second. Therefore, \[\frac{W/s}{Q}=\,0.05\]                 or \[Q\,=\,\,\frac{1000\,{{J}^{-1}}}{0.05}\,\,=20\times \,{{10}^{3}}\,J\,{{s}^{-1}}\]                 \[=20\text{ }kJ\text{ }{{s}^{1}}\]                 \[\therefore \] Heat taken out of the refrigerator per second                 \[=QP=20\text{ }k\text{ }J\text{ }{{s}^{1}}1\text{ }kJ\text{ }{{s}^{1}}\]                 \[=19\text{ }kJ\text{ }{{s}^{1}}.\]