Thermodynamics Questions and Answers – Entropy Principle and its Applications-2

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This set of Thermodynamics Interview Questions and Answers for experienced focuses on “Entropy Principle and its Applications-2”

1. For the flow of electric current through a resistor,
a) at steady state, internal energy of resistor is constant
b) at steady state, temperature of resistor is constant
c) W=Q
d) all of the mentioned
View Answer

Answer: d
Explanation: Internal energy is dependent on temperature and by first law Q=ΔE+W.
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2. When stirring work is supplied to a viscous thermally insulated liquid, temperature of the liquid
a) remains constant
b) increases
c) decreases
d) none of the mentioned
View Answer

Answer: d
Explanation: None.

3. A car uses power of 25 hp for a one hour in a round trip. A thermal efficiency of 35% can be assumed? Find the change in entropy if we assume ambient at 20°C?
a) 554.1 kJ/K
b) 654.1 kJ/K
c) 754.1 kJ/K
d) 854.1 kJ/K
View Answer

Answer: b
Explanation: E = ⌠ W dt = 25 hp × 0.7457 (kW/hp) × 3600 s = 67 113 kJ = η Q
Q = E / η = 67 113 / 0.35 = 191 751 kJ
∆S = Q / T = 191 751 / 293.15 = 654.1 kJ/K.
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4. In a Carnot engine working on ammonia, the high temperature is 60°C and as QH is received, the ammonia changes from saturated liquid to saturated vapor. The ammonia pressure at low temperature is 190 kPa. Find the entropy. thermodynamics-interview-questions-answers-experienced-q4
a) 4.6577 kJ/kg K
b) 5.6577 kJ/kg K
c) 6.6577 kJ/kg K
d) 7.6577 kJ/kg K
View Answer

Answer: a
Explanation: qH = ∫ Tds = T (s2 – s1) = T s(fg) = h2 – h1 = h(fg) = 997.0 kJ/kg
TL = T3 = T4 = Tsat(P) = –20°C
η(cycle) = 1 – (Tl/Th) = 1 – (253.2/333.2) = 0.24
s3 = s2 = sg(60°C) = 4.6577 kJ/kg K.

5. A slab of concrete, 5 × 8 × 0.3 m, is used as a thermal storage mass in a house. The slab cools overnight from 23°C to 18°C in an 18°C house, find the net entropy change associated with this process?
a) 0.4 kJ/K
b) 1.4 kJ/K
c) 2.4 kJ/K
d) 3.4 kJ/K
View Answer

Answer: d
Explanation: V = 5 × 8 × 0.3 = 12 m^3; m = ρV = 2200 × 12 = 26400 kg
V = constant so 1W2 = 0; 1Q2 = mC∆T = 26400 × 0.88(-5) = -116160 kJ
∆S(SYST) = m(s2 – s1) = mC ln(T2/T1) = 26400 × 0.88 ln (291.2/296.2) = -395.5 kJ/K
∆S(SURR) = -1Q2/T0 = +116160/291.2 = +398.9 kJ/K
∆S(NET) = -395.5 + 398.9 = +3.4 kJ/K.
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6. A foundry form box with 25 kg of 200°C hot sand is dropped into a bucket with 50 L water at 15°C. Assuming there is no heat transfer with the surroundings and no boiling away of water, calculate the net entropy change for the process.
a) 2.37 kJ/K
b) 2.47 kJ/K
c) 2.57 kJ/K
d) 2.67 kJ/K
View Answer

Answer: c
Explanation: C.V. Sand and water, constant pressure process
m(sand)∆h(sand) + m(H2O)∆h(H2O) = 0
m(sand)C∆T(sand) + m(H2O)C(H2O)∆T(H2O) = 0
25 × 0.8×(T2 – 200) + (50×10^(-3)/0.001001) × 4.184 × (T2 – 15) = 0
hence T2 = 31.2°C
∆S = 25 × 0.8 ln(304.3/473.15) + 49.95 × 4.184 ln(304.3/288.15)
= 2.57 kJ/K.

7. Calculate the change in entropy if 1 kg of saturated liquid at 30°C is converted into superheated steam at 1 bar and 200°C .
a) 5.3973 kJ/K
b) 6.3973 kJ/K
c) 7.3973 kJ/K
d) none of the mentioned
View Answer

Answer: c
Explanation: si= sf @30 C = 0.4369 kJ/kg.K,
se = sg @1 bar and 200 C = 7.8342 kJ/kg.K
Change in entropy (∆S) = m*( se – si) = 1*(7.8342 – 0.4369)
= 7.3973 kJ/K.
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8. Two kilograms of water at 120°C with a quality of 25% has its temperature raised by 20°C in a constant volume process. What is the new specific entropy?
a) 3.01517 kJ/kg.K
b) 4.01517 kJ/kg.K
c) 5.01517 kJ/kg.K
d) 7.01517 kJ/kg.K
View Answer

Answer: b
Explanation: v1 = vf @120 C + x1*vfg @120 C = 0.00106 + 0.25*0.8908 = 0.22376 m3/kg
v2 = v1 = vf @145 C + x2*vfg @145 C = 0.00108 + x2*0.50777 ∴ x2 = 0.4385
New specific entropy (s2) = sf @145 C + x2*sfg @145 C
= 1.739 +0.4385*5.1908 = 4.01517 kJ/kg.K.

9. A thermal reservoir at 538°C is brought into thermal communication with another thermal reservoir at 260°C, and as a result 1055 kJ of heat is transferred only from the higher to lower temperature reservoir. Determine the change in entropy of the universe due to the exchange of heat between these two thermal reservoirs.
a) 0.378182 kJ/K
b) 0.478182 kJ/K
c) 0.578182 kJ/K
d) 0.678182 kJ/K
View Answer

Answer: d
Explanation: (∆S)System = ∫δQ/T = –1055/(538 + 273.15) + 1055/(260 + 273.15)= 0.678182 kJ/K
(∆S)Surroundings = ∫δQ/T = 0
Change in entropy of the universe ((∆S)Universe)
= (∆S)System + (∆S)Surroundings = 0.678182 kJ/K.

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10. A glass jar is filled with saturated water at 500 kPa of quality 25%, and a tight lid is put on. Now it is cooled to -10°C. What is the mass fraction of solid at this temperature?
a) 99.98%
b) 98.98%
c) 93.98%
d) 95.98%
View Answer

Answer: a
Explanation: Constant volume v1=v2=V/m
from steam table, Psat = 500 kPa and hence Tsat = 151.8°C
v1 = 0.001093 + 0.25*0.3738 = 0.094543
v2 = 0.0010891 + x2*466.756 = v1 = 0.094543
x2 = 0.002 mass fraction vapour
x(solid) = 1- x2 = 0.9998 or 99.98%.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter