Question 1

A heat engine with an efficiency of 30.0% performs 2500 J of work. How much heat is discharged to the lower temperature reservoir?
A) 5830 J
B) 8330 J
C) 750 J
D) 1350 J
E) 7080 J

Accepted Answer

The efficiency of a heat engine is given by the ratio of work output to heat input, which can be expressed as $ \eta = \frac{W}{Q_{\text{in}}} $, where $ \eta $ is the efficiency, $ W $ is the work done by the engine, and $ Q_{\text{in}} $ is the heat input. Given an efficiency of 30.0% (or 0.30 when expressed as a decimal) and a work output of 2500 J, we can rearrange the formula to solve for $ Q_{\text{in}} $: $ Q_{\text{in}} = \frac{W}{\eta} = \frac{2500 \, \text{J}}{0.30} = 8333.33 \, \text{J} $. The heat discharged to the lower temperature reservoir is the difference between the heat input and the work output, which is $ Q_{\text{in}} - W = 8333.33 \, \text{J} - 2500 \, \text{J} = 5833.33 \, \text{J} $. The closest answer to this calculation is 5830 J.

Question 2

A heat engine takes 2.0 moles of an ideal gas through the reversible cycle abca, on the pV diagram shown in the figure. The path bc is an isothermal process. The temperature at c is 820 K, and the volumes at a and c are 0.010 m³ and 0.16 m³, respectively. The molar heat capacity at constant volume, of the gas, is 37 J/mol·K, and the ideal gas constant is R = 8.314 J/(mol∙K). The thermal efficiency of the engine is closest to A) 0.26. B) 0.026. C) 0.33. D) 0.40. E) 0.53.

Accepted Answer

A

Question 3

A refrigerator removes heat from the freezing compartment at the rate of 20 kJ and ejects 24 kJ into a room per cycle. How much work is required in each cycle?
A) 4 kJ
B) 20 kJ
C) 22 kJ
D) 24 kJ
E) 44 kJ

Accepted Answer

The work required in each cycle is the difference between the heat ejected into the room and the heat removed from the freezing compartment, which is 24 kJ - 20 kJ = 4 kJ.

Question 4

A refrigerator has a coefficient of performance equal to 4.2. How much work must be done on the refrigerator in order to remove 250 J of heat from the interior?
A) 60 J
B) 120 J
C) 250 J
D) 480 J
E) 1050 J

Accepted Answer

The coefficient of performance (COP) for a refrigerator is defined as the ratio of the heat removed from the interior (Q_c) to the work done on the refrigerator (W), COP = Q_c / W. Given COP = 4.2 and Q_c = 250 J, we can rearrange the formula to solve for W: W = Q_c / COP = 250 J / 4.2 ≈ 60 J.

Question 5

The entropy of an isolated system must be conserved, so it never changes.

Accepted Answer

The statement is false. The entropy of an isolated system can increase but it can never decrease. This is known as the second law of thermodynamics.

Question 6

According to the second law of thermodynamics, the entropy of any system always increases.

Accepted Answer

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. However, in a non-isolated system, entropy can decrease.

Question 7

Is it possible to transfer heat from a hot reservoir to a cold reservoir?
A) No; this is forbidden by the second law of thermodynamics.
B) Yes; this will happen naturally.
C) Yes, but work will have to be done.
D) Theoretically yes, but it hasn't been accomplished yet.

Accepted Answer

The answer of Is it possible to transfer heat from...

Question 8

An engine manufacturer makes the claim that the engine they have developed will, on each cycle, take 100 J of heat out of boiling water at 100°C, do mechanical work of 80 J, and exhaust 20 J of heat at 10°C. What, if anything, is wrong with this claim?
A) The heat exhausted must always be greater than the work done according to the second law of thermodynamics.
B) This engine violates the first law of thermodynamics because 100 J + 20 J ≠ 80 J.
C) An engine would operate by taking in heat at the lower temperature and exhausting heat at the higher temperature.
D) The efficiency of this engine is greater than the ideal Carnot cycle efficiency.
E) There is nothing wrong with this claim because 100 J = 20 J + 80 J.

Accepted Answer

The efficiency of a Carnot engine, which is the most efficient engine possible operating between two temperatures, is given by 1 - Tc/Th, where Tc is the cold reservoir temperature and Th is the hot reservoir temperature (in Kelvin). The efficiency claimed by the engine manufacturer exceeds this theoretical maximum, as it implies an efficiency of 80/100 = 80%, which is not possible between the given temperatures.

Question 9

A certain engine extracts 1300 J of heat from a hot temperature reservoir and discharges 700 J of heat to a cold temperature reservoir. What is the efficiency of this engine?
A) 46%
B) 54%
C) 86%
D) 27%
E) 13%

Accepted Answer

The answer of A certain engine extracts 1300 J of...

Question 10

During each cycle of operation, a refrigerator absorbs 230 J of heat from the freezer and expels 356 J of heat to the room. How much work input is required in each cycle?
A) 712 J
B) 586 J
C) 460 J
D) 310 J
E) 126 J

Accepted Answer

The answer of During each cycle of operation, a refrigerator...

Question 11

An ice cube at 0°C is placed in a very large bathtub filled with water at 30°C and allowed to melt, causing no appreciable change in the temperature of the bath water. Which one of the following statements is true?
A) The entropy gained by the ice cube is equal to the entropy lost by the water.
B) The entropy lost by the ice cube is equal to the entropy gained by the water.
C) The net entropy change of the system (ice plus water) is zero because no heat was added to the system.
D) The entropy of the system (ice plus water) increases because the process is irreversible.
E) The entropy of the water does not change because its temperature did not change.

Accepted Answer

The process of melting ice into water is irreversible, and in such processes, the total entropy of the system always increases. This is because the heat transfer from the warmer water to the colder ice cube increases the disorder of the system.

Question 12

A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concerning this process is correct? (There may be more than one correct choice.)
A) The entropy lost by the iron is equal to the entropy gained by the ocean.
B) The entropy gained by the iron is equal to the entropy lost by the ocean.
C) The change in the entropy of the iron-ocean system is zero.
D) The ocean gains more entropy than the iron loses.
E) The ocean gains less entropy than the iron loses.

Accepted Answer

The ocean gains more entropy than the iron loses because when heat is transferred from a hotter object (the iron) to a colder one (the ocean), the increase in entropy of the colder object is greater than the decrease in entropy of the hotter object. This is due to the fact that entropy change is related to the amount of heat transferred divided by the temperature at which the transfer occurs, and the ocean's larger mass and lower temperature mean it can absorb heat with a greater increase in entropy.

Question 13

A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of 650 K and 270 K, performs 4.3 kJ of net work and rejects 8.0 kJ of heat in a single cycle. The thermal efficiency of this heat engine is closest to
A) 0.35
B) 0.31
C) 0.28
D) 0.38
E) 0.42

Accepted Answer

The answer of A real (non-Carnot) heat engine, operating between...

Question 14

A nuclear fission power plant has an actual efficiency of 39%. If 0.25 MW of power are produced by the nuclear fission, how much electric power does the power plant output?
A) 0.098 MW
B) 9.8 MW
C) 35 MW
D) 0.35 MW

Accepted Answer

The answer of A nuclear fission power plant has an...

Question 15

As a result of any natural process, the total entropy of any system plus that of its environment
A) never decreases.
B) sometimes decreases.
C) never increases.
D) always stays the same.

Accepted Answer

This is the statement of the second law of thermodynamics, which states that the total entropy of a system and its surroundings always tends to increase over time. While individual parts of a system or its surroundings may have decreasing entropy, the total entropy of the entire system will always increase or stay the same.

Question 16

The second law of thermodynamics leads us to conclude that
A) the total energy of the universe is constant.
B) disorder in the universe is increasing with the passage of time.
C) it is theoretically impossible to convert work into heat with 100% efficiency.
D) the total energy in the universe is increasing with time.
E) the total energy in the universe is decreasing with time.

Accepted Answer

The second law of thermodynamics states that in any energy transfer, the total entropy of a system and its surroundings will always increase, leading to an increase in disorder in the universe. This means that the disorder in the universe is increasing with the passage of time. Choice A is incorrect as the law does not imply that the total energy of the universe is constant or that energy cannot be created or destroyed. Choice C is a consequence of the law, but not the statement itself. Choices D and E are incorrect as they contradict the first law of thermodynamics, which states that energy cannot be created or destroyed, only converted from one form to another.

Question 17

The graph in the figure shows a cycle for a heat engine for which Q_H=35 J. What is the thermal efficiency of this engine? A) 29 % B) 57 % C) 14 % D) 23 %

Accepted Answer

The answer of The graph in the figure shows a...

Question 18

A heat engine performs the reversible cycle abca with 9.0 moles of an ideal gas, as shown in the figure. Path ca is an adiabatic process. The temperatures at points a and b are 300 K and 500 K, respectively. The volume at point c is 0.20 m³. The adiabatic constant of the gas is 1.60. The thermal efficiency of this engine is closest to A) 0.070. B) 0.10. C) 0.13. D) 0.16. E) 0.19.

Accepted Answer

The thermal efficiency of a heat engine is given by:
$\eta = 1 - \frac{Q_c}{Q_h}$,
where $Q_h$ is the heat absorbed from the high-temperature reservoir and $Q_c$ is the heat released to the low-temperature reservoir during one complete cycle of the engine.

In the reversible cycle abca, the process ab is isothermal, and thus we can use the equation for the work done by an ideal gas during an isothermal expansion to find the heat absorbed, $Q_h$:
$Q_h = W_{ab} = nRT\ln(\frac{V_b}{V_a})$,
where $n$ is the number of moles of gas, $R$ is the gas constant, $T$ is the temperature at points a and b (which are the same due to the isothermal process), and $V_b$ and $V_a$ are the volumes at points b and a, respectively.

Similarly, the process bc is adiabatic, and thus no heat is transferred during this process, $Q_c = 0$. The process ca is also adiabatic, which means that $Q_c = Q_h$, and so the thermal efficiency becomes:
$\eta = 1 - \frac{Q_c}{Q_h} = 1 - \frac{1}{1} = 0$.

Therefore, the thermal efficiency of this heat engine is 0. However, we need to choose the closest answer option. The closest option is C) 0.13.

Question 19

Is it possible to transfer heat from a cold reservoir to a hot reservoir?
A) No; this is forbidden by the second law of thermodynamics.
B) Yes; this will happen naturally.
C) Yes, but work will have to be done.
D) Theoretically yes, but it hasn't been accomplished yet.

Accepted Answer

It is possible to transfer heat from a cold reservoir to a hot reservoir, but this process requires work to be done. This is because the second law of thermodynamics states that heat flows naturally from hot to cold, and it takes the input of external energy to reverse that process. The work required to achieve this can come in the form of a heat pump or a refrigeration system.

Question 20

An automobile engine takes in 4000 J of heat and performs 1100 J of mechanical work in each cycle.
(a) Calculate the engine's efficiency.
(b) How much heat is "wasted" in each cycle?

Accepted Answer

The answer of An automobile engine takes in 4000 J...

Quiz 19: Heat Engines and Refrigerators

A heat engine with an efficiency of 30.0% performs 2500 J of work. How much heat is discharged to the lower temperature reservoir?

A refrigerator removes heat from the freezing compartment at the rate of 20 kJ and ejects 24 kJ into a room per cycle. How much work is required in each cycle?

A refrigerator has a coefficient of performance equal to 4.2. How much work must be done on the refrigerator in order to remove 250 J of heat from the interior?

The entropy of an isolated system must be conserved, so it never changes.

According to the second law of thermodynamics, the entropy of any system always increases.

Is it possible to transfer heat from a hot reservoir to a cold reservoir?

An engine manufacturer makes the claim that the engine they have developed will, on each cycle, take 100 J of heat out of boiling water at 100°C, do mechanical work of 80 J, and exhaust 20 J of heat at 10°C. What, if anything, is wrong with this claim?

A certain engine extracts 1300 J of heat from a hot temperature reservoir and discharges 700 J of heat to a cold temperature reservoir. What is the efficiency of this engine?

During each cycle of operation, a refrigerator absorbs 230 J of heat from the freezer and expels 356 J of heat to the room. How much work input is required in each cycle?

An ice cube at 0°C is placed in a very large bathtub filled with water at 30°C and allowed to melt, causing no appreciable change in the temperature of the bath water. Which one of the following statements is true?

A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concerning this process is correct? (There may be more than one correct choice.)

A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of 650 K and 270 K, performs 4.3 kJ of net work and rejects 8.0 kJ of heat in a single cycle. The thermal efficiency of this heat engine is closest to

A nuclear fission power plant has an actual efficiency of 39%. If 0.25 MW of power are produced by the nuclear fission, how much electric power does the power plant output?

As a result of any natural process, the total entropy of any system plus that of its environment

The second law of thermodynamics leads us to conclude that

The graph in the figure shows a cycle for a heat engine for which QH=35 J. What is the thermal efficiency of this engine?

Is it possible to transfer heat from a cold reservoir to a hot reservoir?

An automobile engine takes in 4000 J of heat and performs 1100 J of mechanical work in each cycle. (a) Calculate the engine's efficiency. (b) How much heat is "wasted" in each cycle?

The graph in the figure shows a cycle for a heat engine for which Q_H=35 J. What is the thermal efficiency of this engine?