Verified Solution • 2025–26 Session moderate

QNEET-2016-Ph-1165: NEET 2016 Physics - Class 11 NEET-2016-Ph-1165 Solution: The temperature inside a refrigerator is \(t_2^ C\). The amount …

The temperature inside a refrigerator is t2∘Ct_2^\circ C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be t1−t2t1\frac{t_1-t_2}{t_1} t1t1−t2\frac{t_1}{t_1-t_2} t1+273t1−t2\frac{t_1+273}{t_1-t_2} t2+273t1+273\frac{t_2+273}{t_1+273}

Multiple Choice Options

  • A t1t1−t2{{{t_1}} \over {{t_1} - {t_2}}} t 1 _ _ t 2 _ t 1 _ _
  • B t1+273t1−t2{{{t_1} + 273} \over {{t_1} - {t_2}}} t 1 _ _ t 2 _ t 1 _ + 273 _
  • C t2+273t1−t2{{{t_2} + 273} \over {{t_1} - {t_2}}} t 1 _ _ t 2 _ t 2 _ + 273 _
  • D t1+t2t1+273{{{t_1} + {t_2}} \over {{t_1} + 273}} t 1 _ + 273 t 1 _ + t 2 _ _

Step-by-Step Verified Solution

Temperature inside refrigerator = t 2 �C Room temperature = t 1 �C For refrigerator, H e a t g i v e n t o h i g h t e m p e r a t u r e ( Q 1 ) H e a t t a k e n f r o m l o w e r t e m p e r a t u r e ( Q 2 ) = T 1 T 2 Q 1 Q 2 = t 1 + 273 t 2 + 273 _ Q 1 Q 1 _ W = t 1 + 273 t 2 + 273 _ 1 _ W Q 1 = t 2 + 273 t 1 + 273 _ W Q 1 = t 1 _ t 2 t 1 + 273 The amount of heat delivered to the room for each joule of electrical energy (W = 1 J) Q 1 = t 1 + 273 t 1 _ t 2

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