Question 1

What is the entropy change when 4 kg of water at 0°C is frozen into 4 kg of ice at 0°C?

Accepted Answer

When water freezes, it releases heat to the surroundings, so the change in entropy can be calculated using the formula ΔS = Q/T, where Q is the heat released and T is the temperature at which it is released. Since the water is at 0°C and the ice is also at 0°C, there is no change in temperature and T = 0K. The heat released is given by the heat of fusion of water, which is 334 kJ/kg. Therefore, ΔS = - (334 kJ/kg) / (0K) = - ∞, which is not a physically meaningful answer. However, we know that the entropy change must be negative, since a solid has lower entropy than a liquid. The correct way to calculate the change in entropy is using the equation ΔS = Sice - Swater, where S is the entropy. At 0°C, the entropy of water is 6.023 J/Kg K and the entropy of ice is 1.999 J/Kg K. Therefore, ΔS = (4 kg)(1.999 J/Kg K - 6.023 J/Kg K) = - 17.696 J/K = - 4.9 kJ/K (rounded to one decimal place).

Question 2

The construction of a heat engine operating between temperatures T₁ and T₂, and having an efficiency greater than that of a Carnot engine operating between the same two temperatures, would constitute a violation of

Accepted Answer

The second law of thermodynamics states that it is impossible for any heat engine to have an efficiency greater than that of a Carnot engine operating between two temperatures. Therefore, the construction of such an engine would violate this law.

Question 3

A block of mass m = 0.2 kg slides across a rough horizontal surface with coefficient of kinetic friction μ_k = 0.5. What is the change in entropy after the block has moved a distance of 1 m? The temperature of the block and surrounding is 22°C.

Accepted Answer

The answer of A block of mass m = 0.2...

Question 4

The change in the entropy of the universe due to an operating Carnot engine

Accepted Answer

The Carnot engine is a reversible engine, which means that it operates with maximum theoretical efficiency and causes no net change in the entropy of the system and its surroundings. Therefore, the change in the entropy of the universe due to an operating Carnot engine is zero.

Question 5

A car of mass 2000 kg is traveling at 22.0 m/s on a day when the temperature is 20.0^oC. The driver steps on the brakes and stops the car. By how much does the entropy of the universe increase?

Accepted Answer

The correct answer is A.The change in entropy of the universe is given by the sum of the change in entropy of the car and the change in entropy of the environment. The change in entropy of the car is given by:$$\Delta S_{car} = - \int_{T_i}^{T_f} \frac{dQ}{T}$$where $T_i$ is the initial temperature of the car, $T_f$ is the final temperature of the car, and $dQ$ is the heat transferred from the car to the environment. The change in entropy of the environment is given by:$$\Delta S_{env} = \int_{T_i}^{T_f} \frac{dQ}{T}$$where $T_i$ is the initial temperature of the environment, $T_f$ is the final temperature of the environment, and $dQ$ is the heat transferred from the car to the environment. The change in entropy of the universe is given by:$$\Delta S_{universe} = \Delta S_{car} + \Delta S_{env} = - \int_{T_i}^{T_f} \frac{dQ}{T} + \int_{T_i}^{T_f} \frac{dQ}{T} = 0$$Therefore, the change in entropy of the universe is zero.

Question 6

A container has 0.2 mole of O₂ gas and the gas is heated from 0°C to 100°C. What is the change in entropy of the gas? R = 8.314 J/mol · K.)

Accepted Answer

The change in entropy (ΔS) for an ideal gas when heated at constant volume is given by ΔS = nCvln(T2/T1). For diatomic gases like O2, Cv = (5/2)R. Here, n = 0.2 moles, T1 = 273 K, T2 = 373 K. ΔS = 0.2 * (5/2) * 8.314 * ln(373/273) = 1.82 J/K.

Question 7

Entropy is related to probability. An isolated system moves toward

Accepted Answer

According to the Second Law of Thermodynamics, an isolated system tends to move towards a state of increasing entropy or disorder. Therefore, the state with the highest entropy will be the one with the lowest order and the highest probability of all possible states, which is choice C.

Question 8

A steam power plant with an efficiency of 65% of the maximum thermodynamic efficiency operates between 250^oC and 40^oC. What is the change in the entropy of the universe when this plant does 1.0 kJ of work?

Accepted Answer

The maximum thermodynamic efficiency (Carnot efficiency) is calculated using the temperatures in Kelvin. The actual efficiency is 65% of this value. The change in entropy of the universe is calculated using the heat rejected to the cold reservoir, which can be found from the work done and the actual efficiency.

Question 9

One mole of an ideal gas undergoes a reversible isothermal expansion from a volume of 1 L to a volume of 2 L. The change in entropy of the gas in terms of the universal gas constant R is

Accepted Answer

For an ideal gas undergoing a reversible isothermal process, the change in entropy is given by ΔS = nRln(Vf/Vi), where n is the number of moles of gas, R is the universal gas constant, Vi is the initial volume, and Vf is the final volume. In this case, n = 1 and Vi = 1 L, and Vf = 2 L, so ΔS = R ln(2/1) = R ln2. Therefore, the correct answer is C.

Question 10

Of the four steps in the Carnot cycle,

Accepted Answer

The Carnot cycle consists of two adiabatic processes and two isothermal processes. The adiabatic processes involve no heat transfer, while the isothermal processes occur at constant temperature.

Question 11

A quantity of heat is removed from a hot reservoir at absolute temperature T and then is added to a cold reservoir at absolute temperature T/2. The cold reservoir experiences an entropy increase S. What is the change in the entropy of the universe?

Accepted Answer

The answer of A quantity of heat is removed from...

Question 12

A steam power plant with an efficiency of 65% of the maximum thermodynamic efficiency operates between 250^oC and 40^oC. How much heat is rejected to the cold reservoir in doing 1.0 kJ of work?

Accepted Answer

The answer of A steam power plant with an efficiency...

Question 13

A Carnot engine operating between reservoir temperatures of 340^oC and 40^oC has an efficiency of

Accepted Answer

The answer of A Carnot engine operating between reservoir temperatures...

Question 14

Which of the following statements is true of an isolated system consisting of 15 gas molecules?

Accepted Answer

The second law of thermodynamics states that entropy of a system tends to increase over time until it reaches equilibrium. Since this is an isolated system, there are no external factors that can decrease the entropy. Thus, the entropy of the gas is not likely to decrease. It can either stay the same or increase.

Question 15

When you make ice cubes, the entropy of the water

Accepted Answer

When water freezes to form ice cubes, its molecules become more ordered compared to their arrangement in the liquid state. This increased order means the system's entropy decreases.

Question 16

An electric power plant generates 100 MW of power at an efficiency of 35%. At what rate must water be circulated pass the condenser if the change in water temperature is not to exceed 10^oC?

Accepted Answer

The answer of An electric power plant generates 100 MW...

Question 17

You can vary the efficiency of a Carnot engine by varying the temperature of the cold reservoir while maintaining the hot reservoir at constant temperature. The curve that best represents the efficiency of such an engine as a function of the temperature of the cold reservoir is

Accepted Answer

The answer of  You can vary the efficiency of...

Question 18

Three moles of a gas at T = 250 K expand quasi-statically and adiabatically from an initial volume of 30 L to a final volume of 60 L. The change in entropy of the gas during this expansion is R = 8.314 J/mol·K)

Accepted Answer

In an adiabatic process, there is no heat exchange with the surroundings. The change in entropy for a quasi-static adiabatic process is zero because entropy is a function of heat transfer.

Question 19

Two moles of a gas at T = 350 K expand quasi-statically and isothermally from an initial volume of 20 L to a final volume of 60 L. The change in entropy of the gas during this expansion is R = 8.314 J/mol · K)

Accepted Answer

For an isothermal expansion, the change in entropy can be calculated as:
ΔS = nR ln(Vf/Vi)
where n is the number of moles of the gas and Vf and Vi are the final and initial volumes, respectively. Substituting the given values:
ΔS = 2 mol x 8.314 J/(mol·K) x ln(60 L/20 L) = 18.3 J/K

Therefore, the best choice is B.

Question 20

A thermodynamic system is taken in equilibrium steps from State I to State II. The quantities by which the process can be characterized are 1) internal energy,
2) entropy,
3) temperature,
4) work, and
5) heat.Of these quantities, those that are independent of path are

Accepted Answer

Internal energy, entropy, and temperature are state functions that only depend on the initial and final states of the system, and not on the path taken to get there. On the other hand, work and heat are path-dependent quantities, and can vary depending on the specific process used to go from State I to State II. Therefore, only choices 1, 2, and 3 are independent of path.

Quiz 15: Thermodynamics II

What is the entropy change when 4 kg of water at 0°C is frozen into 4 kg of ice at 0°C?

The construction of a heat engine operating between temperatures T1 and T2, and having an efficiency greater than that of a Carnot engine operating between the same two temperatures, would constitute a violation of

A block of mass m = 0.2 kg slides across a rough horizontal surface with coefficient of kinetic friction μk = 0.5. What is the change in entropy after the block has moved a distance of 1 m? The temperature of the block and surrounding is 22°C.