Thermodynamics

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Question 1

A Carnot engine operating between temperatures T1 and T2 has efficiency 1/6. When T2 is lowered by 62 K, its efficiency increases to 1/3. Then T1 and T2 are, respectively

  • 372 K and 330 K

  • 330 K and 268 K

  • 310 K and 248 K

  • 372 K and 310 K

Solution

D.

372 K and 310 K

The efficiency  is given by,
straight eta subscript 1 space equals space 1 space minus straight T subscript 2 over straight T subscript 1
rightwards double arrow space 1 over 6 space equals space 1 space minus space straight T subscript 2 over straight T subscript 1
rightwards double arrow space straight T subscript 2 over straight T subscript 1 space equals space 5 over 6
straight eta subscript 2 space equals space 1 minus fraction numerator straight T subscript 2 minus 62 over denominator straight T subscript 1 end fraction space space.... space left parenthesis straight i right parenthesis
1 third space equals space 1 minus fraction numerator straight T subscript 2 minus 62 over denominator straight T subscript 1 end fraction space... space left parenthesis ii right parenthesis
On space solving space Eqs space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
straight T subscript 1 space equals space 372 space straight K space and space straight T subscript 2 space equals space 310 space straight K

Question 2

A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

  • 99 J

  • 90 J

  • 1 J

  • 100 J

Solution

B.

90 J

straight W space equals space straight Q subscript 1 space open parentheses straight T subscript 1 over straight T subscript 2 minus 1 close parentheses space space space space straight eta space equals space 1 minus straight T subscript 2 over straight T subscript 1
10 space equals space straight Q subscript 2 space open parentheses 10 over 9 minus 1 close parentheses space space space space 1 over 10 space equals space 1 minus straight T subscript 2 over straight T subscript 1
rightwards double arrow space space straight T subscript 2 over straight T subscript 1 space equals 1 space minus 1 over 10 space equals space 9 over 10
10 space equals space straight Q subscript 2 space open parentheses 1 over 9 close parentheses
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 10 over 9
straight Q subscript 2 space equals space 90 space straight J
Question 3

A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be

  • the efficiency of Carnot engine cannot be made larger than 50%

  • 1200 K

  • 750 K 

  • 600 K

Solution

C.

750 K 

Efficiency
straight eta space equals space 1 minus straight T subscript sink over straight T subscript source
Now comma space 0.4 space equals 1 minus fraction numerator straight T subscript sink over denominator 500 straight K end fraction
rightwards double arrow space straight T subscript sink space equals space 0.6 space straight x space 500 space straight K
space equals space 300 space straight K
Thus comma space 0.6 space equals space 1 minus fraction numerator 300 space straight K over denominator straight T apostrophe subscript source end fraction
rightwards double arrow space straight T to the power of apostrophe subscript source space equals space fraction numerator 300 space straight K over denominator 0.4 end fraction
space equals 750 space straight K

Question 4

A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32 V,the efficiency of the engine is

  • 0.5

  • 0.75

  • 0.99

  • 0.25

Solution

B.

0.75

because space PV to the power of straight gamma space equals space constant
TV to the power of straight gamma minus 1 end exponent space equals space Constant
rightwards double arrow space straight T subscript straight b straight V subscript straight b superscript straight gamma minus 1 end superscript space equals space space straight T subscript straight c straight V subscript straight c superscript straight gamma minus 1 end superscript
rightwards double arrow space straight T subscript straight b over straight V subscript straight c space equals space open parentheses straight V subscript straight c over straight V subscript 0 close parentheses to the power of straight gamma minus 1 end exponent space equals space open parentheses fraction numerator 32 space straight V over denominator straight V end fraction close parentheses to the power of 7 over 5 minus 1 end exponent
space equals space left parenthesis 2 to the power of 5 right parenthesis to the power of 2 over 5 end exponent
straight T subscript straight b space equals space 4 straight T subscript straight c space equals space 4
straight i. straight e. comma space 1 minus straight T subscript straight c over straight T subscript straight b space equals space 1 minus 1 fourth space equals space 3 over 4
Question 5

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies the same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies the same amount of heat. In both the cases body is brought from an initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is:

  • ln2,4ln2

  • ln2,ln2

  • ln2,2ln2

  • 2ln2,8ln2

Solution

B.

ln2,ln2

Since entropy is a state function, therefore a change in entropy in both the processes must be same .

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