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Heat and Thermodynamics question

2016 · Q156
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Heat and Thermodynamics question

2016 · Q156

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
The temperature inside a refrigerator is t2 oC. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
  1. A
    t1t1−t2{{{t_1}} \over {{t_1} - {t_2}}}t1​−t2​t1​​
  2. B
    t1+273t1−t2{{{t_1} + 273} \over {{t_1} - {t_2}}}t1​−t2​t1​+273​
  3. C
    t2+273t1−t2{{{t_2} + 273} \over {{t_1} - {t_2}}}t1​−t2​t2​+273​
  4. D
    t1+t2t1+273{{{t_1} + {t_2}} \over {{t_1} + 273}}t1​+273t1​+t2​​
View written solutionFree

Correct answer: B

Temperature inside refrigerator = t2 °C

Room temperature = t1 °C

For refrigerator,

Heat given to high temperature (Q1)Heat taken from lower temperature (Q2)=T1T2{{{\rm{Heat\,given\,to\,high\,temperature}}\,\left( {{Q_1}} \right)} \over {{\rm{Heat\,taken\,from\,lower \,temperature\, }}\left( {{Q_2}} \right)}} = {{{T_1}} \over {{T_2}}}Heattakenfromlowertemperature(Q2​)Heatgiventohightemperature(Q1​)​=T2​T1​​

Q1Q2=t1+273t2+273{{{Q_1}} \over {{Q_2}}} = {{{t_1} + 273} \over {{t_2} + 273}}Q2​Q1​​=t2​+273t1​+273​

⇒Q1Q1−W=t1+273t2+273 \Rightarrow {{{Q_1}} \over {{Q_1} - W}} = {{{t_1} + 273} \over {{t_2} + 273}}⇒Q1​−WQ1​​=t2​+273t1​+273​

⇒1−WQ1=t2+273t1+273 \Rightarrow 1 - {W \over {{Q_1}}} = {{{t_2} + 273} \over {{t_1} + 273}}⇒1−Q1​W​=t1​+273t2​+273​

⇒WQ1=t1−t2t1+273 \Rightarrow {W \over {{Q_1}}} = {{{t_1} - {t_2}} \over {{t_1} + 273}}⇒Q1​W​=t1​+273t1​−t2​​

The amount of heat delivered to the room for each joule of electrical energy (W = 1 J)

Q1=t1+273t1−t2{Q_1} = {{{t_1} + 273} \over {{t_1} - {t_2}}}Q1​=t1​−t2​t1​+273​

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