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In 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 electrical motor. If the motor is of 1 \mathrm{~kW}power and heat is transferred from -3^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}, find the heat taken out of the refrigerator per second assuming its efficiency is 50 \%  of a perfect engine. 

Option: 1

12 \mathrm{~kJ}


Option: 2

19 \mathrm{~kJ}


Option: 3

10 \mathrm{~kJ}


Option: 4

7 \mathrm{~kJ}


Answers (1)

best_answer

\eta=1-\frac{270}{300}=\frac{1}{10}.  Efficiency of refrigerator =0.5 \eta=\frac{1}{20}

if Q is the heat/s transferred at higher temperature, then 

\mathrm{\frac{W}{Q}=\frac{1}{20}\, or \, Q=20 W=20 \mathrm{~kJ}}

and heat removed from lower temperature \mathrm{=19 \mathrm{~kJ}}.

Posted by

jitender.kumar

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