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Thermodynamics

Answer: B

Only some part of low grade energy is available for conversion.

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12.

A 4 L jug of milk at 25°C is placed in refrigerator where it is cooled down to a temperature of 5°C. Assuming the milk has the property of liquid water, find the entropy generated in the cooling process.

Answer: C

v1 = vf = 0.001003 m3/kg, h = hf = 104.87 kJ/kg; sf = 0.3673 kJ/kg K
h = hf = 20.98 kJ/kg, s = sf = 0.0761 kJ/kg K
P = constant = 101 kPa => 1W2 = mP(v2 – v1);
m = V/v1 = 0.004 / 0.001003 = 3.988 kg
1Q2 = m(h2 − h1) = 3.988 (20.98 – 104.87) = -3.988 × 83.89 = -334.55 kJ
1S2(gen) = m(s2 − s1) − 1Q2/T(refrig)
= 3.988 (0.0761 − 0.3673) − (−334.55 / 278.15) = − 1.1613 + 1.2028
= 0.0415 kJ/K.

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Answer: B

Total entropy increase of the system is the sum of these two entropies.

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Answer: C

As heat is given to a system, its internal energy increases, thus increasing the entropy of the system.

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15.

At ambient temperature, exergy of the fluid is

Answer: D

As the temperature of fluid decreases, its exergy decreases adn when the temperature reaches ambient temperature, its exergy becomes zero.

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16.

In the reversible adiabatic expansion of a gas the increase in disorder due to an increase in volume is compensated by the decrease in disorder due to a decrease in temperature.

Answer: A

This ensures that the disorder number or entropy remains constant.

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17.

The unavailable energy is the product of the lowest temperature of heat rejection and the change of entropy of system during the process of supplying heat.

Answer: A

U.E.=T0*(change in entropy).

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18.

A hot metal piece is cooled rapidly to 25°C, removing 1000 kJ from the metal. Calculate the change of entropy if energy is absorbed by ice.

Answer: B

Ice melting at 0°C; m = 1Q2 /h(fg) = 1000/333.41 = 2.9993 kg
∆S(H2O) = ms(ig) = 2.9993(1.221) = 3.662 kJ/K

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19.

Oxygen gas in a piston cylinder at 300 K, 100 kPa with volume 0.1m^3 is compressed in a reversible adiabatic process to a final temperature of 700 K. Find the final pressure and volume.

Answer: A

Process: Adiabatic 1q2 = 0, Reversible 1s2 gen = 0
Entropy Eq.: s2 – s1 = ∫ dq/T + 1s2 gen = 0
∴s2 = s1 (isentropic compression process)
P2 = P1( T2 / T1)^(k/k-1) = 2015 kPa
V2 = V1( T2 / T1)^(1/1-k) = 0.1 × (700/300)^(1/1−1.393)
= 0.0116 m^3.

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20.

The equation dQ=TdS is true only for a reversible process.

Answer: A

This comes from the second law.

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