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Thermo HW2

The document contains a series of thermodynamics problems and their corresponding answers, focusing on concepts such as Carnot engines, adiabatic processes, and specific heat capacities. It includes calculations related to efficiency, temperature changes, and internal energy changes for various gas processes. The problems are structured to test understanding of thermodynamic principles and provide numeric answers for each scenario.

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0% found this document useful (0 votes)
17 views1 page

Thermo HW2

The document contains a series of thermodynamics problems and their corresponding answers, focusing on concepts such as Carnot engines, adiabatic processes, and specific heat capacities. It includes calculations related to efficiency, temperature changes, and internal energy changes for various gas processes. The problems are structured to test understanding of thermodynamic principles and provide numeric answers for each scenario.

Uploaded by

jane
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Thermodynamics 41

19. Three samples of the same gas A, B and C 25. A Carnot engine whose efficiency is 50% has an
exhaust temperature of 500 K. If the efficiency is to
æ 3ö
ç g = ÷ have initially equal volume. Now the be 60% with the same intake temperature, the exhaust
è 2ø temperature must be (in K)
volume of each sample is double. The process is 26. An engine takes in 5 mole of air at 20°C and 1
adiabatic for A, Isobaric for B and isothermal for atm, and compresses it adiabaticaly to 1/10th of
C. If the finanl pressures are equal for all the
the original volume. Assuming air to be a
three samples, the ratio of their initial pressure is
diatomic ideal gas made up of rigid molecules,
(a) 2 2 : 2 :1 (b) 2 2 :1: 2 the change in its internal energy during this
(c) 2 :1: 2 (d) 2 :1: 2 process comes out to be X kJ. The value of X to
20. A Carnot engine whose low temperature reservoir the nearest integer is ________.
is at 7°C has an efficiency of 50%. It is desired to 27. Starting at temperature 300 K, one mole of an
increase the efficiency to 70%. By how many ideal diatomic gas (g = 1.4) is first compressed
degrees should the temperature of the high
temperature reservoir be increased? V1
adiabatically from volume V1 to V2 = . It is
(a) 840 K (b) 280 K 16
(c) 560 K (d) 373 K then allowed to expand isobarically to volume
2V2. If all the processes are the quasi-static then
Numeric Value Answer the final temperature of the gas (in °K) is (to the
21. An ideal gas at 27ºC is compressed adiabatically nearest integer) ______.
8 28. A Carnot engine operates between two
to of its original volume. The rise in
27 reservoirs of temperatures 900 K and 300 K. The
æ 5ö engine performs 1200 J of work per cycle. The
temperature (in °C) is ç g = ÷
è 3ø heat energy (in J) delivered by the engine to the
22. During an adiabatic process of an ideal gas, if P low temperature reser voir, in a cycle, is
1 _______.
is proportional to 1.5 , then the ratio of specific
V 29. Two Carnot engines A and B are operated in
heat capacities at constant pressure to that at series. The first one, A receives heat at T1
constant volume for the gas is (= 600 K) and rejects to a reservoir at
23. During an adiabatic process, the pressure of a temperature T2. The second engine B receives
gas is found to be proportional to the cube of its
heat rejected by the first engine and in turn,
absolute temperature. The ratio CP/CV for the rejects to a heat reservoir at T3 (= 400 K).
gas is Calculate the temperature T2 (in K) if the work
24. A Carnot freezer takes heat from water at 0°C outputs of the two engines are equal.
inside it and rejects it to the room at a temperature 30. A heat engine is involved with exchange of heat
of 27°C. The latent heat of ice is 336 × 103 J kg– of 1915 J, – 40 J, +125 J and – Q J, during one
1. If 5 kg of water at 0°C is converted into ice at cycle achieving an efficiency of 50.0%. The
0°C by the freezer, then the energy consumed value of Q (in J) is :
(in J) by the freezer is close to :

ANSWER KEY
1 (b) 4 (a) 7 (b) 10 (a) 13 (a) 16 (d) 19 (b) 22 (1.5) 25 (400) 28 (600)
2 (a) 5 (c) 8 (d) 11 (a) 14 (a) 17 (b) 20 (d) 23 (1.5) 26 (46) 29 (500)
5
3 (c) 6 (a) 9 (a) 12 (b) 15 (a) 18 (d) 21 (402) 24 (1.67 × 10 ) 27 (1818) 30 (980)

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