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Mechanical Properties Of Fluids

Question
CBSEENPH11020787
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A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency?

  • 275 K

  • 175 K

  • 250 K

  • 225 K

Solution

C.

250 K

The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied i.e., 
                          straight eta space equals space fraction numerator Work space done over denominator Heat space supplied end fraction space equals straight W over straight Q subscript 1 space equals space fraction numerator straight Q subscript 1 minus straight Q subscript 2 over denominator straight Q subscript 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 minus straight Q subscript 2 over straight Q subscript 1 space equals space 1 minus straight T subscript 2 over straight T subscript 1
Here, straight T subscript 1 is the temperature of source and T2 is the temperature of sink
    As given,             straight eta space equals space 40 percent sign space equals space 40 over 100 space equals 0.4
and                straight T subscript 2 space equals space 300 space straight K
So,        
       0.4 space equals space 1 minus 300 over straight T subscript 1 rightwards double arrow space space space space straight T subscript 1 space equals space fraction numerator 300 over denominator 1 minus 0.4 end fraction equals fraction numerator 300 over denominator 0.6 end fraction space equals 500 space straight K
Let temperature of the source be increased by x K, then efficiency becomes
straight eta apostrophe space equals space 40 percent sign space plus space 50 percent sign space of space straight eta space space space space space equals space 40 over 100 plus 50 over 100 cross times 0.4 space space space space space space equals 0.4 space plus space 0.5 space cross times space 0.4 space space space space space space space equals space 0.6 Hence comma space space space space space 0.6 space equals space 1 minus fraction numerator 300 over denominator 500 plus straight x end fraction rightwards double arrow space space space space space space fraction numerator 300 over denominator 500 plus straight x end fraction space equals space 0.4 rightwards double arrow space space space space space space 500 plus straight x space equals space fraction numerator 300 over denominator 0.4 end fraction space equals space 750
therefore space space space space straight x space equals space 750 minus 500 space equals space 250 space straight K

Some More Questions From Mechanical Properties of Fluids Chapter

Define Dulong and Pettit’s law of specific heat.

Who explained the correct variation of specific heat with temperature in the lower range of temperature?

Do the gases have only one value of specific heat?

What is thermometry?

Define temperature.

When two bodies will be said to be in thermal equilibrium?

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