Physics, asked by Lalitkumath, 5 months ago

Question of thermal physics:-

An Ideal refrigerator works between 27C to 127C temperature. If it expel 120cal heat in 1sec then find avg power in water.

Answers

Answered by Qᴜɪɴɴ
25

Given:-

  • T1 = 127°C
  • T2 = 27°C
  • Expelled heat = Q1 = 120 Cal

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Need to find:-

  • Power = ?

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Concept:-

  • For Refrigerator Heat released = Heat absorbed + work done

  • Power = Work done ÷ Time taken

  • Carnot engine formula

  • When no relation is given, remember to use the formula of Carnot engine for finding necessary values.

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T1 = 127°C

➟ T1 = 127 + 273 K

T1 = 400K

And

T2 = 27°C

➟T2 = 27 + 273

T2 = 300K

We know,

C.O.P.=  \beta  =  \dfrac{Q1}{Q2}  =  \dfrac{T1}{T2}

Substituting their values we get,

 \dfrac{120}{Q2}  =  \dfrac{400}{300}

Q2 =  \dfrac{300 \times 120}{400}

Q2 =  \dfrac{3 \times 120}{4}

\purple{\boxed{\bold{Q2 = 90cal}}}

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We know,

Q1 = Q2 + Work done

➟ 120 Cal = 90 Cal + Work done

➟ Work done = 120 - 90 Cal

➟ Work done = 30 Cal

➟ Work done = 30 × 4.2 Joules

\purple{\boxed{\bold{Work\:done = 126\:Joules}}}

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Power :-

= Work done ÷ Time taken

= 126 Joules ÷ 1 Sec

= 126 Watt

\red{\large{\boxed{\bold{Power=126watt}}}}


Ataraxia: Great! <3
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