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The temperature inside a refrigerator is t_2 ^\circ C and the room temperature is t_1 ^\circ C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

  • Option 1)

    \frac{{t_1 }} {{t_1 - t_2 }}

  • Option 2)

    \frac{{t_1 + 273}} {{t_1 - t_2 }}

  • Option 3)

    \frac{{t_2 + 273}} {{t_1 - t_2 }}

  • Option 4)

    \frac{{t_1 + t_2 }} {{t_1 + 273}}

 

Answers (1)

best_answer

As we learnt in 

Coefficient of performance (beta) -

eta = frac{Q_{2}}{Q_{1}-Q_{2}}= frac{T_{2}}{T_{1}-T_{2}}
 

- wherein

T_{1}> T_{2}

For a perfect refrigerator eta ightarrow infty

 

 Cofficient of performance of a refrigerator is

\beta =\frac{T_2}{T_1-T_2}=\frac{Q_2}{W}

W=1

Q_2=\frac{T_2}{T_1-T_2}=\frac{t_2+273}{t_1-t_2}

Q_1=Q_2 +W=\frac{t_2+273}{t_1-t_2}+1

Q_1=\frac{t_1+273}{t_1-t_2}

Correct option is 2.


Option 1)

\frac{{t_1 }} {{t_1 - t_2 }}

Incorrect

Option 2)

\frac{{t_1 + 273}} {{t_1 - t_2 }}

Correct

Option 3)

\frac{{t_2 + 273}} {{t_1 - t_2 }}

Correct

Option 4)

\frac{{t_1 + t_2 }} {{t_1 + 273}}

Incorrect

Posted by

divya.saini

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