Category: Thermodynamics Objective Questions

Thermodynamics Multiple Choice Questions on “Energy Equation-2”.

1. A 100L rigid tank contains nitrogen at 3 MPa, 900 K. The tank is then cooled to 100 K. What is the heat transfer for this process?
a) −490.7 kJ
b) −590.7 kJ
c) −690.7 kJ
d) −790.7 kJ
Answer: c
Clarification: V = constant hence work = 0 ; Energy Eq: m(u2 – u1) = 1Q2 – 1W2
State 1: v1 = 0.0900 m³/kg => m = V/v1 = 1.111 kg, u1 = 691.7 kJ/kg
State 2: 100 K, v2 = v1 = V/m
interpolating for v (between 200kPa and 400kPa) we get
P2 = 200 + 200 (0.09 – 0.1425)/(0.0681 – 0.1425) = 341 kPa
u2 = 71.7 + (69.3 – 71.7)(0.09 – 0.1425)/(0.0681 – 0.1425) = 70.0 kJ/kg,
1Q2 = m(u2 – u1) = 1.111 (70.0 – 691.7) = −690.7 kJ.

2. A rigid container has 0.75kg water at 1200 kPa, 300°C. Now, the water is cooled to a final pressure of 300 kPa. Find the heat transfer in the process.
a) -2348 kJ
b) -1348 kJ
c) -2148 kJ
d) -1148 kJ
Answer: d
Clarification: Energy Eq.: U2 – U1 = 1Q2 – 1W2 ; V = constant hence work = 0
State 1: 300°C, 1200 kPa => superheated vapor,
thus v = 0.21382 m³/kg, u = 2789.22 kJ/kg
State 2: 300 kPa and v2 = v1 and v2 < vg two-phase we get T2 = Tsat = 133.55°C
x2 = (v2 – vf)/v(fg) = (0.21382 – 0.001073)/0.60475 = 0.35179
u2 = uf + (x2)*u(fg) = 561.13 + x2 1982.43 = 1258.5 kJ/kg
1Q2 = m(u2 – u1) + 1W2 = m(u2 – u1) = 0.75 (1258.5 – 2789.22) = -1148 kJ.

3. A cylinder fitted with a piston contains 2kg of superheated R-134a vapour at 100°C, 350 kPa. The cylinder is then cooled so that R-134a remains at constant pressure till it reaches a quality of 75%. Calculate the heat transfer in this process.
a) -174.6 kJ
b) -274.6 kJ
c) -374.6 kJ
d) -474.6 kJ
Answer: b
Clarification: Energy Eq: m(u2 – u1) = 1Q2 – 1W2
Process: P = constant ⇒ 1W2 =⌡⌠PdV = P∆V = P(V2 – V1) = Pm(v2 – v1)
State 1: h1 = (490.48 + 489.52)/2 = 490 kJ/kg
State 2: h2 = 206.75 + 0.75 ×194.57 = 352.7 kJ/kg (350.9 kPa)
1Q2 = m(u2 – u1) + 1W2 = m(u2 – u1) + Pm(v2 – v1) = m(h2 – h1)
1Q2 = 2 × (352.7 – 490) = -274.6 kJ.

4. Ammonia at 0°C, quality 60% is contained in a 200L tank. The tank and ammonia is now heated to a final pressure of 1 MPa. Determine the heat transfer for the process.
a) 520.75 kJ
b) 620.75 kJ
c) 720.75 kJ
d) 820.75 kJ
Answer: c
Clarification: Energy Eq: m(u2 – u1) = 1Q2 – 1W2
Process: Constant volume hence 1W2 = 0
State 1: two-phase state and v1 = 0.001566 + x1 × 0.28783 = 0.17426 m³/kg
u1 = 179.69 + 0.6 × 1138.3 = 862.67 kJ/kg, m = V/v1 = 0.2/0.17426 = 1.148 kg
State 2: P2 , v2 = v1 superheated vapor ⇒ T2 ≅ 100°C, u2 ≅ 1490.5 kJ/kg
1Q2 = m(u2 – u1) = 1.148(1490.5 – 862.67) = 720.75 kJ.

5. Water in a 150L closed, rigid tank is at 100°C, 90% quality. The tank is cooled to −10°C. Calculate the heat transfer during this process.
a) -163.3 kJ
b) -263.3 kJ
c) -363.3 kJ
d) -463.3 kJ
Answer: b
Clarification: Energy Eq: m(u2 – u1) = 1Q2 – 1W2; Process: V = constant, 1W2 = 0
State 1: Two-phase thus v1 = 0.001044 + 0.9×1.6719 = 1.5057 m³/kg
and u1 = 418.94+0.9×2087.6 = 2297.8 kJ/kg
State 2: T2, v2 = v1 ⇒ mix of saturated solid + vapour
v2 = 1.5057 = 0.0010891 + x2 × 466.7 => x2 = 0.003224
u2 = -354.09 + 0.003224 × 2715.5 = -345.34 kJ/kg; m = V/v1 = 0.15/1.5057
= 0.09962 kg
1Q2 = m(u2 – u1) = 0.09962(-345.34 – 2297.8) = -263.3 kJ.

6. A cylinder with constant volume of 0.1 L contains water at critical point. It then cools down to room temperature of 20°C. Calculate the heat transfer from the water.
a) -61.7 kJ
b) -71.7 kJ
c) -81.7 kJ
d) -91.7 kJ
Answer: a
Clarification: Energy Eq: m(u2 – u1) = 1Q2 – 1W2
Process: Constant volume ⇒ v2 = v1 hence work done is zero.
State 1: v1 = vc = 0.003155 m³/kg, u1 = 2029.6 kJ/kg and m = V/v1 = 0.0317 kg
State 2: T2, v2 = v1 = 0.001002 + x2 × 57.79
x2 = 3.7×10^(-5), u2 = 83.95 + x2 × 2319 = 84.04 kJ/kg
1Q2 = m(u2 – u1) = 0.0317(84.04 – 2029.6) = -61.7 kJ.

7. A constant pressure piston-cylinder contains 0.2 kg water as saturated vapour at 400 kPa. It is now cooled so that the water occupies half the original volume. Find the heat transfer in the process.
a) –203.9 kJ
b) –233.9 kJ
c) –223.9 kJ
d) –213.9 kJ
Answer: d
Clarification: Energy Eq: m(u2 – u1) = 1Q2 – 1W2
and P = constant => 1W2 = Pm(v2 – v1)
thus 1Q2 = m(u2 – u1) + 1W2 = m(u2 – u1) + Pm(v2 – v1) = m(h2 – h1)
State 1: v1 = 0.46246 m³/kg; h1 = 2738.53 kJ/kg
State 2: v2 = v1 / 2 = 0.23123 = vf + x v(fg)
x2 = (v2 – vf) / v(fg) = (0.23123 – 0.001084) / 0.46138 = 0.4988
h2 = hf + (x2)*h(fg) = 604.73 + 0.4988 × 2133.81 = 1669.07 kJ/kg
1Q2 = 0.2 (1669.07 – 2738.53) = –213.9 kJ.

8. 2kg water at 120°C with a quality of 25% has its temperature raised 20°C in a constant volume process. What is the heat transfer in the process?
a) 877.8 kJ
b) 887.8 kJ
c) 897.8 kJ
d) 907.8 kJ
Answer: a
Clarification: Energy Eq.: m (u2 − u1 ) = 1Q2 − 1W2 and V = constant thus work is zero
State 1: T, x1 and v1 = vf + (x1)*v(fg) = 0.00106 + 0.25 × 0.8908 = 0.22376 m³/kg
u1 = uf + (x1)*u(fg) = 503.48 + 0.25 × 2025.76 = 1009.92 kJ/kg
State 2: T2, v2 = v1 < vg2 = 0.50885 m³/kg so two-phase
x2 = (v2 – vf2)/v(fg2) = (0.22376 – 0.00108)/0.50777 = 0.43855
u2 = u(f2) + (x2)*u(fg2) = 588.72 + x2 ×1961.3 = 1448.84 kJ/kg
From the energy equation, 1Q2 = m(u2 − u1) = 2 ( 1448.84 – 1009.92 ) = 877.8 kJ.

9. A 25 kg mass moving with 25 m/s is brought to a complete stop with a constant deceleration over a period of 5 seconds by a brake system. The brake energy is absorbed by 0.5kg water initially at 100 kPa, 20°C. Assume that the mass is at constant P and T. Find the energy the brake removes from the mass assuming P = C.
a) 7.6125 kJ
b) 7.7125 kJ
c) 7.8125 kJ
d) 7.9125 kJ
Answer: c
Clarification: E2 – E1= ∆E = 0.5 mV² = 0.5 × 25 × 25^(2)/1000 = 7.8125 kJ.

10. An insulated cylinder fitted with a piston contains R-12 at 25°C with a quality of 90% and V=45 L. The piston moves and the R-12 expands until it exists as saturated vapour. During this, R-12 does 7kJ of work against the piston. Determine the final temperature, assuming that the process is adiabatic.
a) -5°C
b) -15°C
c) -25°C
d) -35°C
Answer: b
Clarification: Energy Eq.: m(u2 − u1) = 1Q2 – 1W2
State 1: (T, x) => v1 = 0.000763 + 0.9 × 0.02609 = 0.024244 m³/kg
m = V1/v1 = 0.045/0.024244 = 1.856 kg and u1 = 59.21 + 0.9 × 121.03 = 168.137 kJ/kg
Q = 0 = m(u2 – u1) + 1W2 = 1.856 × (u2 – 168.137) + 7.0
=> u2 = 164.365 kJ/kg = ug at T2 and T2 comes out to be -15°C.

11. A reactor filled with water having volume 1 m³ is at 360°C, 20 MPa and placed inside a containment room which is well insulated and initially evacuated. Due to a failure, the reactor ruptures and water fills the room. Find the minimum room volume so that the final pressure does not exceed 200 kPa.
a) 257.7 m³
b) 267.7 m³
c) 277.7 m³
d) 287.7 m³
Answer: d
Clarification: Mass: m2 = m1 = V(reactor)/v1 = 1/0.001823 = 548.5 kg
Energy: m(u2 – u1) = 1Q2 – 1W2 = 0 – 0 = 0 hence u2 = u1
State 1: v1 = 0.001823 m³/kg; u1 = 1702.8 kJ/kg which is also equal to u2
State 2: P2 = 200 kPa, u2 Two-phase
x2 = (u2 – uf)/u(fg) = (1702.8 – 504.47)/2025.02 = 0.59176
v2 = 0.001061 + 0.59176 × 0.88467 = 0.52457 m³/kg
V2 = m2 v2 = 548.5 ×0.52457 = 287.7 m³.

12. A piston-cylinder has the piston loaded with outside atmospheric pressure and piston mass to a pressure of 150 kPa. It contains water at −2°C, which is heated until the water becomes saturated vapour. Find the specific work for the process.
a) 163.7 kJ/kg
b) 173.7 kJ/kg
c) 183.7 kJ/kg
d) 193.7 kJ/kg
Answer: b
Clarification: Energy Eq. per unit mass: u2 – u1 = 1q2 – 1w2
Process: P = constant = P1, => work = P1(v2 – v1)
State 1: T1 , P1 => saturated solid; v1 = 1.09×10^(-3) m³/kg, u1 = -337.62 kJ/kg
State 2: x = 1, P2 = P1 = 150 kPa; v2 = vg(P2) = 1.1593 m³/kg,
T2 = 111.4°C ; u2 = 2519.7 kJ/kg
work = P1(v2 -v1) = 150[1.1593 -1.09×10^(-3)] = 173.7 kJ/kg.

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Thermodynamics Multiple Choice Questions on “Efficiencies in Steam Power Plant “.

1. In a gas power cycle,
a) the working fluid is gas
b) it does not undergo phase change
c) engines which operate on gas cycle can be cyclic or non-cyclic
d) all of the mentioned
Answer: d
Clarification: These are basic facts about gas power cycles.

2. A Carnot cycle consists of
a) two reversible isotherms and two reversible isobars
b) two reversible isotherms and two reversible adiabatics
c) two reversible isotherms and two reversible isochores
d) two reversible isobars and two reversible adiabatics
Answer: b
Clarification: This is what a Carnot cycle means.

3. The efficiency of a Carnot cycle is given by
a) 1 – (T1/T2)
b) heat supplied / net work
c) net work / heat supplied
d) all of the mentioned
Answer: c
Clarification: Here T1 is the temperature of heat addition and T2 is the temperature of heat rejection.

4. For a Carnot cycle, the large back work is a big disadvantage.
a) true
b) false
Answer: a
Clarification: This is true for both Carnot gas cycle and Carnot vapour cycle.

5. The Stirling cycle consists of
a) two reversible isotherms and two reversible isobars
b) two reversible isotherms and two reversible adiabatics
c) two reversible isotherms and two reversible isochores
d) two reversible isobars and two reversible adiabatics
Answer: c
Clarification: The Stirling cycle consists of these processes.

6. The efficiency of Stirling cycle is ____ the efficiency of Carnot cycle.
a) more than
b) less than
c) equal to
d) none of the mentioned
Answer: b
Clarification: The reason being in Stirling cycle, heat transfers are constant volume processes.

7. The efficiency of regenerative Stirling cycle is ____ the efficiency of Carnot cycle.
a) more than
b) less than
c) equal to
d) none of the mentioned
Answer: c
Clarification: This is because of the regenerative arrangement used.

8. The Ericsson cycle consists of
a) two reversible isotherms and two reversible isobars
b) two reversible isotherms and two reversible adiabatics
c) two reversible isotherms and two reversible isochores
d) two reversible isobars and two reversible adiabatics
Answer: a
Clarification: The Ericsson cycle is made up of these processes.

9. The efficiency of Ericsson cycle is ____ the efficiency of Carnot cycle.
a) more than
b) less than
c) equal to
d) none of the mentioned
Answer: b
Clarification: In Ericsson cycle, part of heat is transferred at constant temperature and part at constant pressure.

10. The regenerative, Stirling and Ericsson cycles have same efficiency as Carnot cycle.
a) true
b) false
Answer: a
Clarification: But the back work is less compared to Carnot cycle.

Thermodynamics Multiple Choice Questions on “One Dimensional Steady Isentropic Flow and Choking-2”.

1. The discharge is maximum when Mach number is
a) 0
b) 1
c) 2
d) infinity
Answer: b
Clarification: This comes when we differentiate discharge with respect to M.

2. Which of the following relation is correct?
a) (dA)/A = (M + 1)*(dV/V)
b) (dA)/A = (M – 1)*(dV/V)
c) (dA)/A = (M² + 1)*(dV/V)
d) (dA)/A = (M² – 1)*(dV/V)
Answer: d
Clarification: Taking isentropic flow and using continuity equation. we get this relation.

3. Where do we have the value of Mach number as 1 ?
a) at the start of flow
b) at the throat
c) at the end of flow
d) everywhere along the nozzle or diffuser
Answer: b
Clarification: At throat we have area as constant which is required for M=1.

4. Which of the following statement is true about velocity of gas when it is subsonic?
a) it is subsonic before throat
b) becomes sonic at throat
c) becomes supersonic till its exit
d) all of the mentioned
Answer: d
Clarification: This is the case when convergent-divergent duct acts as nozzle.

5. When are properties at throat termed as critical?
a) when M=1
b) discharge is maximum
c) nozzle is choked
d) all of the mentioned
Answer: d
Clarification: For a choked nozzle, we have critical properties at throat.

6. Which of the following relation is correct?(asterik * indicates critical properties)
a) T*/T = 2/ɣ
b) T*/T = 2/(ɣ-1)
c) T*/T = 2/(ɣ+1)
d) T*/T = 2/(ɣ-2)
Answer: c
Clarification: This is the critical temperature ratio.

7. For diatomic gases, critical pressure ratio is equal to
a) 0.528
b) 0.628
c) 0.728
d) 0.828
Answer: a
Clarification: Putting ɣ=1.4 in critical pressure ratio relation we get this.

8. Which of the following statement is true?
a) M is proportional to only velocity
b) M tends towards infinity at low speeds
c) both of the mentioned
d) none of the mentioned
Answer: d
Clarification: Mach number is not proportional to velocity alone and tends towards infinity at very high speeds.

9. Which of the following is correct?
a) when M<1, M*>1 and when M>1, M*<1
b) when M=0, M*=0 and when M=1, M*=1
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: When M<1, M*<1 and when M>1, M*>1.

10. At the choking limit, nozzle passes the ____ mass flow.
a) maximum
b) minimum
c) constant
d) all of the mentioned
Answer: a
Clarification: This occurs at the throat.

11. The condition of supersonic flow through throat with adiabatic conditions is called the design pressure ratio of nozzle.
a) true
b) false
Answer: b
Clarification: The condition of supersonic flow through throat with isentropic conditions is called the design pressure ratio of nozzle.

12. Which of the following statement is true?
a) shocks occur when flow is supersonic
b) after the shock, flow becomes subsonic
c) rest of the diverging portion acts as diffuser
d) all of the mentioned
Answer: d
Clarification: Shocks is a highly irreversible phenomenon.

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Thermodynamics Multiple Choice Questions on “Displacement Work and Indicator Diagram”.

1. Macroscopic properties p and V are significant only for
a) equilibrium states
b) non-equilibrium states
c) depends on the state
d) none of the mentioned
Answer: a
Clarification: This is true for at any intermediate point in the travel of piston.

2. In a cylinder, infinitesimal amount of work done by the gas on piston is given by
a) F*dl
b) p*a*dl
c) p*dV
d) all of the mentioned
Answer: d
Clarification: F=p*a and work equals force multiplied by displacement.

3. For a constant pressure process, work done is
a) zero
b) p*(V2-V1)
c) p1*V1*ln(V2/V1)
d) none of the mentioned
Answer: b
Clarification: Work done in a process is given by the area under p-dV graph.

4. For a constant volume process, work done is
a) zero
b) p*(V2-V1)
c) p1*V1*ln(V2/V1)
d) none of the mentioned
Answer: a
Clarification: Work done in a process is zero since volume remains constant.

5. For a process in which pV=C, work done is
a) zero
b) p*(V2-V1)
c) p1*V1*ln(V2/V1)
d) none of the mentioned
Answer: c
Clarification: Work done is given by integral of (p*dV) from V1 to V2.

6. The area of the indicator diagram represents the magnitude of the
a) net heat transfer by the system
b) net work done on the system
c) net work done by the system
d) none of the mentioned
Answer: c
Clarification: It is the work done in one engine cycle.

7. For Pm=(Ad/Ld)*K, which of the following is true?
a) Pm=mean effective pressure
b) Ad and Ld are area of diagram and length of diagram respectively
c) K id the indicator spring constant
d) all of the mentioned
Answer: d
Clarification: This formula comes from the indicator diagram.

8. For a two-stroke engine, work done in one minute is given by
a) PmALN
b) PmALN/2
c) 2PmALN
d) none of the mentioned
Answer: a
Clarification: In a two-stroke cycle, the engine cycle is completed in two strokes of the piston.

9. The power available at crankshaft is always _____ indicated power.
a) more
b) less
c) equal
d) none of the mentioned
Answer: b
Clarification: Due to friction.

10. Mechanical efficiency of engine is given by
a) IP/BP
b) 1/(BP*IP)
c) (BP*IP)
d) BP/IP
Answer: d
Clarification: Brake power is less than indicated power and mechanical efficiency is given by BP/IP.

11. An engine is said to be double-acting, if the working fluid is made to work on both sides of the piston.
a) true
b) false
Answer: a
Clarification: Such an engine develops twice the amount of work developed in a single-acting engine.

here is complete set of 1000+ Multiple Choice Questions and Answers.

Thermodynamics Questions and Answers for experienced on “Kelvin-Planck Statement and Clausius’ Statement of Second Law”.

1. According to Kelvin-Planck statement, it is ____ for a heat engine to produce net work in a complete cycle if it exchanges heat only with bodies at ____
a) impossible, single fixed temperature
b) possible, changing temperature
c) impossible, changing temperature
d) possible, single fixed temperature
Answer: a
Clarification: This is the basic definition of Kelvin-Planck statement.

2. If heat rejected from the system Q2 is zero, then
a) net work=Q1 and efficiency=1.00
b) heat is exchanged only with one reservoir
c) it violates the Kelvin-Planck statement
d) all of the mentioned
Answer: d
Clarification: Such a heat engine is called a perpetual motion machine of the second kind(PMM2).

3. A PMM2 is possible.
a) true
b) false
Answer: b
Clarification: A PMM2 is impossible because it violates the Kelvin-Planck statement.

4. A heat engine has to exchange heat with ___ energy reservoir at ___ different temperatures to produce net work in a complete cycle.
a) one, one
b) one, two
c) two, two
d) none of the mentioned
Answer: c
Clarification: This is required to produce power.

5. The second law is not a deduction of the first law.
a) true
b) false
Answer: a
Clarification: The first law is a separate law of nature.

6. The continual operation of a machine that creates its own energy and thus violates the first law is called
a) PMM2
b) PMM1
c) PMM0
d) none of the mentioned
Answer: b
Clarification: This is a basic fact about PMM1.

7. Which of the following is true?
a) heat always from a high temperature body to a low temperature body
b) heat always from a low temperature body to a high temperature body
c) heat can flow from both low to high and high to low temperature body
d) none of the mentioned
Answer: a
Clarification: The reverse process never occurs spontaneously.

8. According to Clausius statement
a) it is impossible to construct a device than can transfer heat from a cooler body to a hotter body without any effect
b) it is impossible to construct a device than can transfer heat from a hotter body to a cooler body without any effect
c) it is possible to construct a device than can transfer heat from a cooler body to a hotter body without any effect
d) none of the mentioned
Answer: a
Clarification: To transfer heat from a cooler body to a hotter body, some work must be expended.

9. If the second law were not true
a) a ship could be driven by extracting heat from the ocean
b) run a power plant by extracting heat from the air
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Both of the above possibilities do not violate the first law but do violate the second law.

10. The operation of a machine that utilizes the internal energy of only one TER, thus violating second law is called
a) PMM0
b) PMM1
c) PMM2
d) none of the mentioned
Answer: b
Clarification: PMM2 violates the second law.

Thermodynamics for Experienced,

Thermodynamics Multiple Choice Questions on “Quality of Energy”.

1. A hot gas flowing through a pipeline can be considered as a
a) reversible process
b) irreversible process
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: The process given here is irreversible.

2. For an infinitesimal reversible process at constant pressure,
a) dS=m*dT/T
b) dS=Cp*dT/T
c) dS=m*dT/T
d) dS=m*Cp*dT/T
Answer: d
Clarification: Here m is the mass of gas flowing, Cp is its specific heat, and T is the gas temperature.

3. Adequate insulation must be provided for high temperature fluids.
a) true
b) false
Answer: a
Clarification: This is done to prevent heat loss which would be high at high temperatures.

4. The loss of exergy is more when,
a) the heat loss occurs at a higher temperature
b) the heat loss occurs at a lower temperature
c) depends on the process
d) none of the mentioned
Answer: a
Clarification: Higher the temperature, more will be the exergy loss.

5. The exergy of a fluid at a higher temperature is ___ than that at a lower temperature and ___ as the
temperature decreases.
a) less, increases
b) more, increases
c) more, decreases
d) less, decreases
Answer: c
Clarification: Higher is the temperature, higher will be the exergy and it decreases as the temperature decreases.

6. The second law affixes a quality to energy of a system at any state.
a) true
b) false
Answer: a
Clarification: For example we can say that the quality of energy of a gas at say 1000K is superior to that
at say 300K.

7. The available energy of a system ___ as its temperature or pressure decreases and approaches that of
the surroundings.
a) increases
b) decreases
c) remains constant
d) none of the mentioned
Answer: b
Clarification: As temperature decreases, exergy decreases.

8. At ambient temperature, exergy of the fluid is
a) neagtive
b) positive
c) infinity
d) zero
Answer: d
Clarification: As the temperature of fluid decreases, its exergy decreases adn when the temperature reaches ambient temperature, its exergy becomes zero.

9. Which of the following is true?
a) the first law states that the energy is always conserved quantity-wise
b) the second law states that the energy always degrades quality-wise
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Energy is always conserved according to first law but its quality is degraded.

10. If the first law is the conservation of energy, then the second law is called the law of degradation of
energy.
a) true
b) false
Answer: a
Clarification: Though energy is always conserved but its quality is always degraded.

11. When a gas is throttled adiabatically from a high to a low pressure,
a) the enthalpy remains same
b) there is degradation of energy
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is because of the fact that energy always degrades quality-wise.

12. The ___ in entropy in an irreversible change is a measure of the extent to which energy ___ in that
change.
a) decrease, degrades
b) increase, degrades
c) increase, increases
d) decrease, increases
Answer: b
Clarification: Also, to get maximum work from a system, changes must be performed in a reversible manner.

13. If two bodies were allowed to reach thermal equilibrium, one by heat conduction and other by operating a Carnot engine between them and extracting work, the final equilibrium temperatures would
be different.
a) this is because of the lower value of the total internal energy
b) this is because of the higher value of the total internal energy
c) both of the mentioned
d) none of the mentioned
Answer: a
Clarification: Work is done at the expense of internal energy.

14. Adequate insulation is not that important for low temperature fluids.
a) true
b) false
Answer: a
Clarification: This is because the loss of available energy from such fluids would be low.