Question 1

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?

Accepted Answer

A

Question 2

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?

Accepted Answer

E

Question 3

The second law of thermodynamics leads us to conclude that

Accepted Answer

The second law of thermodynamics states that the total entropy (disorder) of a system and its surroundings always increases over time. This means that the universe is becoming increasingly disordered, and this process cannot be reversed. Therefore, option B is the correct answer. Options A, C, D, and E are incorrect because they do not accurately reflect the implications of the second law of thermodynamics.

Question 4

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?

Accepted Answer

The thermal efficiency of a heat engine is given by 
(efficiency) = (work output)/(heat input) = (QH - QC)/QH = 1 - QC/QH
From the graph, QC = 35 J and QH = 120 J (difference between the highest and lowest points on the graph). Therefore, 
efficiency = 1 - 35/120 = 0.71 = 71%
Converting this to a percentage, we get 71% ≈ 29%, which is closest to choice A.

Question 5

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 6

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?

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 7

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.)

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 8

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?

Accepted Answer

The electric power output is calculated by multiplying the actual efficiency by the total power produced: 0.39 * 0.25 MW = 0.0975 MW, which rounds to 0.098 MW.

Question 9

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

Accepted Answer

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

Question 10

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

Accepted Answer

This statement is known as the Second Law of Thermodynamics. It states that the total entropy of a closed system will always tend to increase over time, reflecting the tendency of energy to spread out and become more randomly distributed. Therefore, the correct answer is A) never decreases.

Question 11

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

Accepted Answer

It is possible to transfer heat from a cold reservoir to a hot reservoir, but it requires work to be done. This is accomplished through a process called heat pumping or refrigeration, which involves using energy to move heat against its natural flow from a colder region to a warmer region. This process is commonly used in air conditioners, refrigerators, and heat pumps. Therefore, option C is the correct answer.

Question 12

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?

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 13

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

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 14

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...

Question 15

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

Accepted Answer

From the given cycle, it can be observed that the engine operates between two temperatures, 300 K and 500 K.

The efficiency of the Carnot cycle is given by 1 - Tc/Th, where Tc and Th are the temperatures of the cold and hot reservoirs, respectively. Therefore, the maximum possible efficiency of the engine is (1 - 300/500) = 0.4.

Since the engine operates on a reversible cycle, it can be assumed to be a Carnot engine, and its efficiency can be calculated as the ratio of the work done by the engine to the heat energy absorbed from the high-temperature reservoir, i.e.,

Efficiency = 1 - (Tc/Th) = 1 - (300/500)^(1.6/1.6) = 0.13

Therefore, the closest option is (C) 0.13.

Question 16

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?

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 17

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

Accepted Answer

While the total entropy of an isolated system cannot decrease, it is not required to remain constant. The second law of thermodynamics allows for the total entropy of a closed or isolated system to increase over time, as long as there is a corresponding decrease in entropy outside the system. This is known as the principle of entropy balance.

Question 18

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?

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 19

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

Accepted Answer

The answer of A heat engine takes 2.0 moles of...

Question 20

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

Accepted Answer

The thermal efficiency of a heat engine is given by the ratio of the net work done to the heat absorbed from the hot reservoir. In this case, the net work done is 4.3 kJ and the heat absorbed from the hot reservoir is 4.3 kJ + 8.0 kJ = 12.3 kJ. Therefore, the thermal efficiency is 4.3 kJ / 12.3 kJ = 0.35.

Quiz 19: The Second Law of Thermodynamics