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THERMODYNAMICS
EFFICIENCY OF CARNOT ENGINE
Efficiency = (1 -
𝑇2
𝑇1
) X 100%
𝑇2 = temperature of sink
𝑇1 = temperature of source
Source thulo sink sano
@ sano by thulo
1) Efficiency of carnot engine working between 27o C and 127 o C is
a) 50%
b) 100%
c) 25%
d) 75%
Ans: c
2) In a carnot engine, the temperature of heat sink is 27 o C and that of the source is
327o C. The efficiency of the engine is
a) 50%
b) 100%
c) 25%
d) 75%
Ans: a
3) A carnot engine has the same efficiency between 800K to 500K and
xK to 600K, the value of x is
a) 1000k
b) 960K
c) 846K
d) 754k
Ans: b
Efficiency =
π‘Š
Q1
X 100%
Where Q1 is heat taken from source
Q2 = heat released into sink
Q1 = Q2 + Work done
1) A carnot engine operated at 800K and 200K. If the engine takes 8J of
heat then what is the amount of work done by the engine
a) 6J
b) 4J
c) 10J
d) 2J
Ans: a
3) The efficiency of a carnot’s engine is 33%. If the system gives out heat at 100 o C it takes heat at
a) 86 o C
b) 283 o C
c) 6 o C
d) 8 o C
Ans: b
4) The temperature of source is 500K with source energy 200J, what is the temperature of sink with sink energy
100J
a) 500K
b) 300K
c) 250K
d) 125K
Ans: c
dQ = dU + dW
dQ = +ve = heat given to system
= -ve = heat taken from system
dU = change in internal energy
= only depends on temperature
dW = Pβˆ†V
= gas expands = volume increase = βˆ†V +ve = dW +ve
= gas contracts = volume decrease = βˆ†V -ve = dW –ve
Work is done on system = +ve
Work is done by system = -ve
1) If 1500J of heat is supplied to a system and 1000J of work is done. What is the change in internal energy?
a) 500J
b) 5300J
c) 4300J
d) 6000J
Ans: a
2) If 1500J of heat is supplied to a system and 1000J of work is done. What is the change in internal energy?
a) 500J
b) 5300J
c) 4300J
d) 6000J
Ans: b
1) Internal energy of an ideal gas depends on
a) Volume and temperature
b) Pressure and temperature
c) Volume only
d) Temperature only
Ans: d
Isothermal process
Temperature constant
Change in temperature = zero
Change in internal energy = zero
PV = constant i.e. P 1 V1 = P2 V 2
Adiabatic process
Heat constant
Change in heat = 0
PVΟ’ = constant
1) The pressure and volume are changing but the temperature is constant in the process
a) Adiabatic
b) Isothermal
c) Isobaric
d) Isochoric
Ans: b
2) In an isothermal condition
a) Heat given by surrounding is equal to work done
b) Heat given by surrounding is equal to internal energy
c) There is no heat loss to the surrounding
d) All
Ans: a
3) In an isothermal process
a) Pressure remains constant
b) Thermal energy remains constant
c) Volume remains constant
d) Temperature remains constant
Ans: d
Adiabatic process (dQ = 0)
dQ = dU + dW
So, dU = -dW
1) If a gas is allowed to expand adiabatically against external pressure
a) Its temperature remains constant
b) Pressure remains constant
c) There is increase in internal energy
d) There is decrease in internal energy
Ans: d
2) What happens in adiabatic process
a) Volume remains constant
b) Pressure remains constant
c) Temperature remains constant
d) The system is insulated from the surrounding
Ans: d
3) When a gas undergoes adiabatic expansion, its internal energy
a) Increases
b) Decreases
c) Remains same
d) None
Ans: b
4) During adiabatic compression of 5 moles of gas, 250J of work was done, the change in internal energy will be
a) 150J
b) 250J
c) -150J
d) -250J
Ans: d
1) Adiabatic elasticity is ……. times adiabatic elasticity
a) Ο’
b) 2Ο’
c) 3Ο’
d) 4Ο’
Ans: a
ISOBARIC PROCESS = pressure constant
ISOCHORIC PROCESS = volume constant
1) In isochoric process ………. is constant
a) Pressure
b) Volume
c) Both pressure and volume
d) None of above
Ans: b
2) In isobaric process ………. is constant
a) Pressure
b) Volume
c) Both pressure and volume
d) None of above
Ans: a
CALORIMETRY
Specific heat capacity of water = 4200 (SI)
= 1 (CGS)
Specific heat capacity of ice = 2100 (SI)
= 0.5 (CGS)
Latent heat of ice = 80cal/gm
Latent heat of steam = 540cal/gm
1) The amount of heat required to convert 1gm of ice at -100 oC to steam at 100 oC is
a) 770 cal
b) 676 cal
c) 736 cal
d) 540 cal
Ans: a
2) 10gm of ice at -10oC is converted into steam at 100oC the amount of heat required in calories is
a) 725
b) 7250
c) 350
d) 3000
Ans: b
3) Heat required to convert 1gm of ice at 0oC into steam at 100oC is
a) 100cal
b) 0.01kcal
c) 720cal
d) 200cal
Ans: c
1) Water is used as coolant due to
a) Low specific heat
b) High specific heat
c) Universal solvent
d) Anamolous expansion
Ans: b
2) As compared to a person with white skin another person with dark skin will experience
a) Less heat and more cold
b) More heat and more cold
c) More heat and less cold
d) Less heat and less cold
Ans: b
3) A cooking utensil should have (VVI)
a) Low specific heat and high thermal conductivity
b) Low specific heat and low thermal conductivity
c) High specific heat and high thermal conductivity
d) High specific heat and low thermal conductivity
Ans: a
ANAMOLOUS EXPANSION OF WATER
1) At what condition the following substance will expand on heating
and cooling
a) Water at 0oC
b) Water at 4oC
c) Ice at 0oC
d) Semiconductor
Ans: b
1) When liquid changes into vapour, on increasing pressure the boiling
point of the liquid
a) Increases
b) decreases
c) Cannot be predicted
d) May increase or decrease
Ans: a
THERMOMETRY
πΆβˆ’0
100
=
πΉβˆ’32
180
=
πΎβˆ’273
100
πΆβˆ’πΏπΉπ‘ƒ
π‘ˆπΉπ‘ƒ βˆ’πΏπΉπ‘ƒ
=
πΉβˆ’πΏπΉπ‘ƒ
π‘ˆπΉπ‘ƒβˆ’πΏπΉπ‘ƒ
=
πΎβˆ’πΏπΉπ‘ƒ
π‘ˆπΉπ‘ƒβˆ’πΏπΉπ‘ƒ
1) A thermometry reads the body temperature in Fahrenheit scale gives reading
140o F. what is the reading in Celsius scale
a) 60o C
b) 140o C
c) 95.6o C
d) 75o C
Ans: a
2) The temperature of a body recorded by Celsius thermometer is -50 o C its
temperature recorded by kelvin scale is
a) 223K
b) 23K
c) 33K
d) 333K
Ans: a
3) A faulty thermometer with lower fixed point and upper fixed point 5 o C and 95 o
C respectively. What is the correct temperature when the reading gives 35
a) 60o C
b) 33.3o C
c) 44o C
d) 66o C
Ans: b
4) A faulty thermometer reads melting point of ice as -10o C. It reads 60o C in place
of 50 o C. What is the temperature of boiling point of water on this scale
a) 130o C
b) 100o C
c) 13o C
d) 30o C
Ans: a
THERMAL EXPANSION
𝑙 = 𝑙o 1 + π›Όβˆ†πœƒ
A = 𝐴o 1 + π›½βˆ†πœƒ
V = Vo 1 + π›Ύβˆ†πœƒ
𝛼 :𝛽 :𝛾 = 1:2:3
1) A zinc rod has a length 1m at 0oC. Find its length at 50oC. The linear expansivity of zinc
is 26 x 10-6
a) 1.0013m
b) 4.013m
c) 3.13m
d) 2.3m
Ans: a
2) The value of 𝛼 :𝛽 :𝛾 =
a) 1:2:3
b) 4:2:3
c) 5:2:3
d) 3:2:1
Ans: a
SIMPLE PENDULUM
In summer pendulum clock runs slower
In winter pendulum clock runs faster
βˆ†π’•
𝒕
=
𝟏
𝟐
π›Όβˆ†πœƒ
1) The pendulum clock is made of brass. If the clock keeps correct
time at 20 oC, how many seconds per day will it lose at 35 oC
(𝛼 for brass = 2 x 10-5)
a) 12.3 sec
b) 2.3 sec
c) 22.3 sec
d) 32.3 sec
Ans: a
THANKYOU

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Thermodynamics || Physics || B.Optom

  • 2. EFFICIENCY OF CARNOT ENGINE Efficiency = (1 - 𝑇2 𝑇1 ) X 100% 𝑇2 = temperature of sink 𝑇1 = temperature of source Source thulo sink sano @ sano by thulo
  • 3. 1) Efficiency of carnot engine working between 27o C and 127 o C is a) 50% b) 100% c) 25% d) 75% Ans: c 2) In a carnot engine, the temperature of heat sink is 27 o C and that of the source is 327o C. The efficiency of the engine is a) 50% b) 100% c) 25% d) 75% Ans: a
  • 4. 3) A carnot engine has the same efficiency between 800K to 500K and xK to 600K, the value of x is a) 1000k b) 960K c) 846K d) 754k Ans: b
  • 5. Efficiency = π‘Š Q1 X 100% Where Q1 is heat taken from source Q2 = heat released into sink Q1 = Q2 + Work done
  • 6. 1) A carnot engine operated at 800K and 200K. If the engine takes 8J of heat then what is the amount of work done by the engine a) 6J b) 4J c) 10J d) 2J Ans: a
  • 7. 3) The efficiency of a carnot’s engine is 33%. If the system gives out heat at 100 o C it takes heat at a) 86 o C b) 283 o C c) 6 o C d) 8 o C Ans: b 4) The temperature of source is 500K with source energy 200J, what is the temperature of sink with sink energy 100J a) 500K b) 300K c) 250K d) 125K Ans: c
  • 8. dQ = dU + dW dQ = +ve = heat given to system = -ve = heat taken from system dU = change in internal energy = only depends on temperature dW = Pβˆ†V = gas expands = volume increase = βˆ†V +ve = dW +ve = gas contracts = volume decrease = βˆ†V -ve = dW –ve Work is done on system = +ve Work is done by system = -ve
  • 9. 1) If 1500J of heat is supplied to a system and 1000J of work is done. What is the change in internal energy? a) 500J b) 5300J c) 4300J d) 6000J Ans: a 2) If 1500J of heat is supplied to a system and 1000J of work is done. What is the change in internal energy? a) 500J b) 5300J c) 4300J d) 6000J Ans: b
  • 10. 1) Internal energy of an ideal gas depends on a) Volume and temperature b) Pressure and temperature c) Volume only d) Temperature only Ans: d
  • 11. Isothermal process Temperature constant Change in temperature = zero Change in internal energy = zero PV = constant i.e. P 1 V1 = P2 V 2 Adiabatic process Heat constant Change in heat = 0 PVΟ’ = constant
  • 12. 1) The pressure and volume are changing but the temperature is constant in the process a) Adiabatic b) Isothermal c) Isobaric d) Isochoric Ans: b 2) In an isothermal condition a) Heat given by surrounding is equal to work done b) Heat given by surrounding is equal to internal energy c) There is no heat loss to the surrounding d) All Ans: a
  • 13. 3) In an isothermal process a) Pressure remains constant b) Thermal energy remains constant c) Volume remains constant d) Temperature remains constant Ans: d
  • 14. Adiabatic process (dQ = 0) dQ = dU + dW So, dU = -dW
  • 15. 1) If a gas is allowed to expand adiabatically against external pressure a) Its temperature remains constant b) Pressure remains constant c) There is increase in internal energy d) There is decrease in internal energy Ans: d 2) What happens in adiabatic process a) Volume remains constant b) Pressure remains constant c) Temperature remains constant d) The system is insulated from the surrounding Ans: d
  • 16. 3) When a gas undergoes adiabatic expansion, its internal energy a) Increases b) Decreases c) Remains same d) None Ans: b 4) During adiabatic compression of 5 moles of gas, 250J of work was done, the change in internal energy will be a) 150J b) 250J c) -150J d) -250J Ans: d
  • 17. 1) Adiabatic elasticity is ……. times adiabatic elasticity a) Ο’ b) 2Ο’ c) 3Ο’ d) 4Ο’ Ans: a
  • 18. ISOBARIC PROCESS = pressure constant ISOCHORIC PROCESS = volume constant
  • 19. 1) In isochoric process ………. is constant a) Pressure b) Volume c) Both pressure and volume d) None of above Ans: b 2) In isobaric process ………. is constant a) Pressure b) Volume c) Both pressure and volume d) None of above Ans: a
  • 21. Specific heat capacity of water = 4200 (SI) = 1 (CGS) Specific heat capacity of ice = 2100 (SI) = 0.5 (CGS) Latent heat of ice = 80cal/gm Latent heat of steam = 540cal/gm
  • 22. 1) The amount of heat required to convert 1gm of ice at -100 oC to steam at 100 oC is a) 770 cal b) 676 cal c) 736 cal d) 540 cal Ans: a 2) 10gm of ice at -10oC is converted into steam at 100oC the amount of heat required in calories is a) 725 b) 7250 c) 350 d) 3000 Ans: b
  • 23. 3) Heat required to convert 1gm of ice at 0oC into steam at 100oC is a) 100cal b) 0.01kcal c) 720cal d) 200cal Ans: c
  • 24. 1) Water is used as coolant due to a) Low specific heat b) High specific heat c) Universal solvent d) Anamolous expansion Ans: b 2) As compared to a person with white skin another person with dark skin will experience a) Less heat and more cold b) More heat and more cold c) More heat and less cold d) Less heat and less cold Ans: b
  • 25. 3) A cooking utensil should have (VVI) a) Low specific heat and high thermal conductivity b) Low specific heat and low thermal conductivity c) High specific heat and high thermal conductivity d) High specific heat and low thermal conductivity Ans: a
  • 26. ANAMOLOUS EXPANSION OF WATER 1) At what condition the following substance will expand on heating and cooling a) Water at 0oC b) Water at 4oC c) Ice at 0oC d) Semiconductor Ans: b
  • 27. 1) When liquid changes into vapour, on increasing pressure the boiling point of the liquid a) Increases b) decreases c) Cannot be predicted d) May increase or decrease Ans: a
  • 30. 1) A thermometry reads the body temperature in Fahrenheit scale gives reading 140o F. what is the reading in Celsius scale a) 60o C b) 140o C c) 95.6o C d) 75o C Ans: a 2) The temperature of a body recorded by Celsius thermometer is -50 o C its temperature recorded by kelvin scale is a) 223K b) 23K c) 33K d) 333K Ans: a
  • 31. 3) A faulty thermometer with lower fixed point and upper fixed point 5 o C and 95 o C respectively. What is the correct temperature when the reading gives 35 a) 60o C b) 33.3o C c) 44o C d) 66o C Ans: b 4) A faulty thermometer reads melting point of ice as -10o C. It reads 60o C in place of 50 o C. What is the temperature of boiling point of water on this scale a) 130o C b) 100o C c) 13o C d) 30o C Ans: a
  • 33. 𝑙 = 𝑙o 1 + π›Όβˆ†πœƒ A = 𝐴o 1 + π›½βˆ†πœƒ V = Vo 1 + π›Ύβˆ†πœƒ 𝛼 :𝛽 :𝛾 = 1:2:3
  • 34. 1) A zinc rod has a length 1m at 0oC. Find its length at 50oC. The linear expansivity of zinc is 26 x 10-6 a) 1.0013m b) 4.013m c) 3.13m d) 2.3m Ans: a 2) The value of 𝛼 :𝛽 :𝛾 = a) 1:2:3 b) 4:2:3 c) 5:2:3 d) 3:2:1 Ans: a
  • 35. SIMPLE PENDULUM In summer pendulum clock runs slower In winter pendulum clock runs faster βˆ†π’• 𝒕 = 𝟏 𝟐 π›Όβˆ†πœƒ
  • 36. 1) The pendulum clock is made of brass. If the clock keeps correct time at 20 oC, how many seconds per day will it lose at 35 oC (𝛼 for brass = 2 x 10-5) a) 12.3 sec b) 2.3 sec c) 22.3 sec d) 32.3 sec Ans: a