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CBSE Class 11-science Answered

A  carnot  engine  has  an  efficiency  of 1/10 as  heat engine, is used  as  refrigeraator.  if  the  work  done on  the  system is  10joule , What  is the  amount   of  energy  absorbed from  the  reservoir  at  lower  temperature  ?  
Asked by PARDEEP | 31 Mar, 2018, 04:01: PM
answered-by-expert Expert Answer
carnot Engine efficiency  begin mathsize 12px style eta space equals space 1 minus Q subscript 2 over Q subscript 1 end style
where Q2 is heat energy rejected at low temperature reservoir and Q1 heat energy absorbed at high temperature reservoir.
 
hence begin mathsize 12px style Q subscript 2 over Q subscript 1 space equals space 1 space minus space eta end style
coefficient of performence of refrigerator α = begin mathsize 12px style fraction numerator Q subscript 2 over denominator Q subscript 1 minus Q subscript 2 end fraction space equals space fraction numerator begin display style bevelled Q subscript 2 over Q subscript 1 end style over denominator 1 minus begin display style bevelled Q subscript 2 over Q subscript 1 end style end fraction space equals space fraction numerator 1 minus eta over denominator eta end fraction space equals space fraction numerator 0.9 over denominator 0.1 end fraction space equals space 9 end style
In refrigerator the workdone W = Q1-Q2 and begin mathsize 12px style alpha space equals space Q subscript 2 over W end style
Hence Q2 = α×W = 9×10 = 90 Joules
Answered by Thiyagarajan K | 01 Apr, 2018, 03:39: PM
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