Question 1

A Carnot engine operates between two reservoirs at 550 K and 250 K.It transfers 300 J to the reservoir at low temperature.The amount of work done is

Accepted Answer

The efficiency of a Carnot engine is given by $ \eta = 1 - \frac{T_{C}}{T_{H}} $, where $ T_{C} $ is the temperature of the cold reservoir and $ T_{H} $ is the temperature of the hot reservoir. Here, $ T_{C} = 250 $ K and $ T_{H} = 550 $ K, so $ \eta = 1 - \frac{250}{550} = \frac{300}{550} $. The work done $ W $ is related to the heat rejected $ Q_{C} $ and the efficiency by $ W = Q_{H} - Q_{C} $, where $ Q_{H} $ is the heat absorbed from the hot reservoir. Since $ \eta = \frac{W}{Q_{H}} $, we can rearrange to find $ Q_{H} = \frac{W}{\eta} $. Given $ Q_{C} = 300 $ J, and knowing that $ Q_{H} = Q_{C} + W $, we can solve for $ W $ using the efficiency equation. Rearranging and solving gives $ W = \eta \cdot Q_{H} = \frac{300}{550} \cdot (Q_{C} + W) $. Solving this equation for $ W $ gives $ W = 360 $ J.

Question 2

A water cooled engine produces 1000 W of power.The engine's thermal efficiency is 36.4% and the rate of water flow is 100.If water enters the engine block at 15S1U1P1$\circ$S1S1P0C,what is the exit temperature of water?

Accepted Answer

The answer of A water cooled engine produces 1000 W...

Question 3

You have just bought a Carnot refrigerator to keep your food cold.If you intend to run your refrigerator between the cold temperature reservoir at 2.0S1U1P1$\circ$S1S1P0C (the inside of the box)and the high temperature reservoir at 26S1U1P1$\circ$S1S1P0C (your kitchen temperature)how much work will be needed to draw 15 J of heat from the cold reservoir?

Accepted Answer

The work needed for a Carnot refrigerator can be calculated using the formula: $W = Q_C \left( \frac{T_H}{T_C} - 1 \right)$, where $Q_C$ is the heat extracted from the cold reservoir, $T_H$ and $T_C$ are the temperatures of the hot and cold reservoirs respectively, in Kelvin. First, convert the temperatures from Celsius to Kelvin: $T_C = 10 + 273.15 = 283.15 K$ and $T_H = 26 + 273.15 = 299.15 K$. Then, plug the values into the formula: $W = 15 \left( \frac{299.15}{283.15} - 1 \right) \approx 0.847 J$. However, since this specific answer is not among the options and considering the nature of the question, it seems there might have been a misunderstanding in the calculation or the interpretation of the provided options. The correct approach involves using the Carnot efficiency or performance coefficient formula correctly with the given temperatures, ensuring they are in Kelvin, and applying the correct physics principles. Re-evaluation of the calculation steps or clarification of the options might be necessary.

Question 4

A 145-g snowball,initially at 0S1U1P1$\circ$S1S1P0C,melts in your hands becoming 145 g of water at 20S1U1P1$\circ$S1S1P0C.What is the change in entropy of the snow/water associated with this event? Remember the latent heat of fusion for water is 334 kJ/kg and the specific heat of liquid water is 4.19 kJ/(kgK).

Accepted Answer

The change in entropy is calculated by summing the entropy change during melting and the entropy change during heating. For melting: ΔS_melting = mL/T = (0.145 kg)(334 kJ/kg) / 273 K. For heating: ΔS_heating = mc ln(T2/T1) = (0.145 kg)(4.19 kJ/kgK) ln(293/273). Adding these gives approximately 0.22 kJ/K.

Question 5

You have just bought a Carnot refrigerator to keep your frozen vegetables frozen.If you intend to run your refrigerator between the cold temperature reservoir at -5S1U1P1$\circ$S1S1P0C (the inside of the box)and the high temperature reservoir at 26S1U1P1$\circ$S1S1P0C (your kitchen temperature)how much work will be needed to draw 15 J of heat from the cold reservoir?

Accepted Answer

To find the work needed for a Carnot refrigerator, we use the formula: $W = Q_C \left( \frac{T_H}{T_C} - 1 \right)$, where $Q_C$ is the heat extracted from the cold reservoir, $T_H$ and $T_C$ are the temperatures of the hot and cold reservoirs, respectively, in Kelvin. First, convert temperatures from Celsius to Kelvin: $T_C = -5 + 273.15 = 268.15 K$ and $T_H = 26 + 273.15 = 299.15 K$. Then, $W = 15 \left( \frac{299.15}{268.15} - 1 \right) \approx 1.7 J$.

Question 6

500-g of water at 100S1U1P1$\circ$S1S1P0C is changed into steam at 100S1U1P1$\circ$S1S1P0C at constant pressure.The change in entropy of the water is

Accepted Answer

The change in entropy ($\Delta S$) when water is converted to steam at constant temperature and pressure can be calculated using the formula $\Delta S = \frac{Q}{T}$, where $Q$ is the heat absorbed and $T$ is the absolute temperature in Kelvin. The heat absorbed ($Q$) can be found from the latent heat of vaporization of water, which is approximately 2260 kJ/kg at 100°C. Since we have 500 g (or 0.5 kg) of water, $Q = 0.5 \times 2260 = 1130$ kJ. The temperature $T$ must be in Kelvin, so we convert 100°C to Kelvin by adding 273.15, giving us 373.15 K. Thus, $\Delta S = \frac{1130}{373.15}$ kJ/K, which is approximately 3.03 kJ/K.

Question 7

A 16-gram lead bullet slams into a concrete wall and melts on impact.If the temperature of the speeding bullet was 25S1U1P1$\circ$S1S1P0C and if the specific heat of lead is 0.13 kJ/(kgK)and its latent heat of fusion is 23 kJ/kg,what is the change in entropy of the bullet during this process? The melting point of lead is 601K.

Accepted Answer

The change in entropy can be calculated using the formula ΔS = Q/T, where Q is the heat absorbed or released during the process and T is the temperature. In this case, the bullet absorbs heat equal to the product of its mass, the latent heat of fusion, and the fraction of the bullet that melts, which is (16 g) × (23 kJ/kg) × (1/16) = 23 kJ. The temperature of the bullet is given in Fahrenheit, so it needs to be converted to Kelvin: (25 − 32) × 5/9 + 273 = 268 K. Therefore, ΔS = 23 kJ / 268 K = 2.08 J/K, which is closest to choice A.

Question 8

One mole of a monatomic ideal gas at a pressure of 3.0 atm and a volume of 30 L is isothermally expanded to a pressure of 1.0 atm and a volume of 90 L.Afterwards it is compressed at a constant pressure until its volume is 30 L and then it's pressure is increased at a constant volume back to the original volume of 30 L as shown in the PV diagram.What is the efficiency of this heat engine cycle?

Accepted Answer

The answer of One mole of a monatomic ideal gas...

Question 9

An 800-kg car traveling at 25 m/s crashes into a wall on a hot day (27S1U1P1$\circ$S1S1P0C).What is the total change in entropy?

Accepted Answer

We need to calculate the change in entropy using the formula ΔS = Q/T, where Q is the heat energy absorbed by the car (due to the collision) and T is the temperature of the car and the environment after the collision. The temperature change can be assumed to be negligible.

To calculate Q, we can use the formula Q = 0.5 * mv^2, where m is the mass of the car and v is the velocity of the car just before the collision. 
Q = 0.5 * 800 kg * (25 m/s)^2 = 250,000 J

Therefore, ΔS = 250,000 J / (27°C + 273) K = 830 J/K.

The correct answer is C.

Question 10

What size heat pump with a coefficient of performance of 4.0 is needed for a home that loses heat energy at a rate of 12 kW on the coldest day of the year?

Accepted Answer

A heat pump with a coefficient of performance (COP) of 4.0 can deliver 4 times as much heat energy as the electrical energy it consumes. Therefore, to provide 12 kW of heat energy, it needs 12 kW / 4 = 3 kW of electrical input.

Question 11

Your Carnot refrigerator has a coefficient of performance of 5.4.Assume that 15.0 J of heat are removed from the cold refrigerator compartment in each engine cycle.What is the heat expelled into the kitchen for each cycle?

Accepted Answer

The coefficient of performance (COP) for a refrigerator is defined as the ratio of the heat removed from the cold reservoir (Q_c) to the work input (W), COP = Q_c / W. Given that the COP is 5.4 and Q_c is 15.0 J, we can find the work input as W = Q_c / COP = 15.0 J / 5.4 = 2.8 J. The heat expelled into the kitchen (Q_h) is the sum of the heat removed from the cold compartment and the work input, Q_h = Q_c + W = 15.0 J + 2.8 J = 17.8 J.

Question 12

Burning of fuel transfers 5 * 10⁵ W of power into the engine of a 1500-kg vehicle;suppose the maximum speed the vehicle can achieve 5 seconds after starting from rest is 32.6 m/s.Determine the efficiency of the engine.

Accepted Answer

Efficiency (η) can be calculated using the formula: η = (Work output / Energy input) x 100% The work output is equal to the kinetic energy gained by the vehicle: Work output = (1/2)mv^2 = (1/2)(1500 kg)(32.6 m/s)^2 = 795450 J The energy input is equal to the power delivered to the engine multiplied by the time it operates: Energy input = (Power)(Time) = (5 F1F1F1S1F1F1F10 10⁵ W)(5 s) = 2.5x10^8 J Therefore, the efficiency is: η = (795450 J / 2.5x10^8 J) x 100% = 0.31818... x 100% = 32% (rounded to 2 significant figures) The closest to this value is choice C, 32%.

Question 13

Burning of fuel transfers power into the engine of a 2000-kg vehicle;suppose the engine has an efficiency of 25% and the maximum speed the vehicle can achieve 5 seconds after starting from rest is 32.6 m/s.Determine the power the fuel transfers to the engine.

Accepted Answer

The answer of Burning of fuel transfers power into the...

Question 14

An ideal Carnot engine takes in 160 kcal of heat at 100S1U1P1$\circ$S1S1P0C and exhausts some of this at 35S1U1P1$\circ$S1S1P0C.How much work must have been done by the engine?

Accepted Answer

The work done by a Carnot engine is given by W = Qh * (1 - Tc/Th), where Qh is the heat absorbed, and Tc and Th are the absolute temperatures of the cold and hot reservoirs, respectively. Converting temperatures to Kelvin and using the given heat absorbed, the work done is approximately 116.7 kJ.

Question 15

You toss two dice;the total number of ways you can get 12 is

Accepted Answer

The only way to get a total of 12 with two dice is by rolling a 6 on both dice.

Question 16

If an engine takes in 600 J of heat and puts out 200 J of heat,what is its efficiency?

Accepted Answer

The answer of If an engine takes in 600 J...

Question 17

A heat pump is operating between two reservoirs at -10S1U1P1$\circ$S1S1P0C and 50S1U1P1$\circ$S1S1P0C.The maximum possible coefficient of performance for the heat pump is

Accepted Answer

The maximum possible coefficient of performance (COP) for a heat pump is given by the formula: COP = T_hot / (T_hot - T_cold), where T_hot and T_cold are the temperatures of the hot and cold reservoirs, respectively, in Kelvin. First, convert the temperatures from Celsius and Fahrenheit to Kelvin. -10°C = 263.15K and 50°C = 323.15K. Plugging these into the formula gives COP = 323.15 / (323.15 - 263.15) = 323.15 / 60 = 5.3858, which rounds to 5.38, making option D correct.

Question 18

Burning of fuel transfers 4 * 10⁵ W of power into the engine of a 2000-kg vehicle.Suppose the engine has an efficiency of 25% and the vehicle reaches a maximum speed of 32.6 m/s.Assuming the vehicle starts from rest,how long does it take the vehicle to reach this speed?

Accepted Answer

The answer of Burning of fuel transfers 4 * 10⁵...

Question 19

A system undergoes a reversible adiabatic process.The entropy change of the system is

Accepted Answer

In a reversible adiabatic process, there is no heat exchange with the surroundings and the system is perfectly insulated. Therefore, the entropy change of the system is zero (i.e., ΔS=0). This is because entropy is a measure of the system's disorder, and in a reversible process, the system returns to its initial state, implying that there is no change in its disorder.

Question 20

A bartender mixes 0.1 liter of alcohol which is at room temperature (= 300 K)with 0.3 liter of soda water out of the refrigerator (temperature = 280 K).In this process,

Accepted Answer

When two substances at different temperatures are mixed, heat flows from the warmer one to the colder one until they reach thermal equilibrium. Therefore, in this case, heat will flow from the alcohol (at 300 K) to the soda water (at 280 K). As a result, the entropy of the water will increase due to the absorption of heat, while the entropy of the alcohol will decrease due to its loss of heat. The net change in entropy will depend on the specific heat capacities of the substances, but since the question does not provide that information, we cannot determine the exact values. However, we can conclude that the entropy of the system as a whole (alcohol + water) has increased, as is expected in any spontaneous process.

Quiz 20: The Second Law of Thermodynamics

A Carnot engine operates between two reservoirs at 550 K and 250 K.It transfers 300 J to the reservoir at low temperature.The amount of work done is

A water cooled engine produces 1000 W of power.The engine's thermal efficiency is 36.4% and the rate of water flow is 100.If water enters the engine block at 15^$\circ$C,what is the exit temperature of water?

A 145-g snowball,initially at 0^$\circ$C,melts in your hands becoming 145 g of water at 20^$\circ$C.What is the change in entropy of the snow/water associated with this event? Remember the latent heat of fusion for water is 334 kJ/kg and the specific heat of liquid water is 4.19 kJ/(kgK) .

500-g of water at 100^$\circ$C is changed into steam at 100^$\circ$C at constant pressure.The change in entropy of the water is

An 800-kg car traveling at 25 m/s crashes into a wall on a hot day (27^$\circ$C) .What is the total change in entropy?

What size heat pump with a coefficient of performance of 4.0 is needed for a home that loses heat energy at a rate of 12 kW on the coldest day of the year?

Your Carnot refrigerator has a coefficient of performance of 5.4.Assume that 15.0 J of heat are removed from the cold refrigerator compartment in each engine cycle.What is the heat expelled into the kitchen for each cycle?

Burning of fuel transfers 5 * 10⁵ W of power into the engine of a 1500-kg vehicle;suppose the maximum speed the vehicle can achieve 5 seconds after starting from rest is 32.6 m/s.Determine the efficiency of the engine.

Burning of fuel transfers power into the engine of a 2000-kg vehicle;suppose the engine has an efficiency of 25% and the maximum speed the vehicle can achieve 5 seconds after starting from rest is 32.6 m/s.Determine the power the fuel transfers to the engine.

An ideal Carnot engine takes in 160 kcal of heat at 100^$\circ$C and exhausts some of this at 35^$\circ$C.How much work must have been done by the engine?

You toss two dice;the total number of ways you can get 12 is

If an engine takes in 600 J of heat and puts out 200 J of heat,what is its efficiency?

A heat pump is operating between two reservoirs at -10^$\circ$C and 50^$\circ$C.The maximum possible coefficient of performance for the heat pump is

Burning of fuel transfers 4 * 10⁵ W of power into the engine of a 2000-kg vehicle.Suppose the engine has an efficiency of 25% and the vehicle reaches a maximum speed of 32.6 m/s.Assuming the vehicle starts from rest,how long does it take the vehicle to reach this speed?

A system undergoes a reversible adiabatic process.The entropy change of the system is

A bartender mixes 0.1 liter of alcohol which is at room temperature (= 300 K) with 0.3 liter of soda water out of the refrigerator (temperature = 280 K) .In this process,

Quiz 20: The Second Law of Thermodynamics

A Carnot engine operates between two reservoirs at 550 K and 250 K.It transfers 300 J to the reservoir at low temperature.The amount of work done is

A water cooled engine produces 1000 W of power.The engine's thermal efficiency is 36.4% and the rate of water flow is 100.If water enters the engine block at 15 ∘\circ∘ C,what is the exit temperature of water?

500-g of water at 100 ∘\circ∘ C is changed into steam at 100 ∘\circ∘ C at constant pressure.The change in entropy of the water is

An 800-kg car traveling at 25 m/s crashes into a wall on a hot day (27 ∘\circ∘ C) .What is the total change in entropy?

What size heat pump with a coefficient of performance of 4.0 is needed for a home that loses heat energy at a rate of 12 kW on the coldest day of the year?

Your Carnot refrigerator has a coefficient of performance of 5.4.Assume that 15.0 J of heat are removed from the cold refrigerator compartment in each engine cycle.What is the heat expelled into the kitchen for each cycle?

Burning of fuel transfers 5 * 105 W of power into the engine of a 1500-kg vehicle;suppose the maximum speed the vehicle can achieve 5 seconds after starting from rest is 32.6 m/s.Determine the efficiency of the engine.

Burning of fuel transfers power into the engine of a 2000-kg vehicle;suppose the engine has an efficiency of 25% and the maximum speed the vehicle can achieve 5 seconds after starting from rest is 32.6 m/s.Determine the power the fuel transfers to the engine.

An ideal Carnot engine takes in 160 kcal of heat at 100 ∘\circ∘ C and exhausts some of this at 35 ∘\circ∘ C.How much work must have been done by the engine?

You toss two dice;the total number of ways you can get 12 is

If an engine takes in 600 J of heat and puts out 200 J of heat,what is its efficiency?

A heat pump is operating between two reservoirs at -10 ∘\circ∘ C and 50 ∘\circ∘ C.The maximum possible coefficient of performance for the heat pump is

Burning of fuel transfers 4 * 105 W of power into the engine of a 2000-kg vehicle.Suppose the engine has an efficiency of 25% and the vehicle reaches a maximum speed of 32.6 m/s.Assuming the vehicle starts from rest,how long does it take the vehicle to reach this speed?

A system undergoes a reversible adiabatic process.The entropy change of the system is

A bartender mixes 0.1 liter of alcohol which is at room temperature (= 300 K) with 0.3 liter of soda water out of the refrigerator (temperature = 280 K) .In this process,

A water cooled engine produces 1000 W of power.The engine's thermal efficiency is 36.4% and the rate of water flow is 100.If water enters the engine block at 15^$\circ$C,what is the exit temperature of water?

500-g of water at 100^$\circ$C is changed into steam at 100^$\circ$C at constant pressure.The change in entropy of the water is

An 800-kg car traveling at 25 m/s crashes into a wall on a hot day (27^$\circ$C) .What is the total change in entropy?

Burning of fuel transfers 5 * 10⁵ W of power into the engine of a 1500-kg vehicle;suppose the maximum speed the vehicle can achieve 5 seconds after starting from rest is 32.6 m/s.Determine the efficiency of the engine.

An ideal Carnot engine takes in 160 kcal of heat at 100^$\circ$C and exhausts some of this at 35^$\circ$C.How much work must have been done by the engine?

A heat pump is operating between two reservoirs at -10^$\circ$C and 50^$\circ$C.The maximum possible coefficient of performance for the heat pump is

Burning of fuel transfers 4 * 10⁵ W of power into the engine of a 2000-kg vehicle.Suppose the engine has an efficiency of 25% and the vehicle reaches a maximum speed of 32.6 m/s.Assuming the vehicle starts from rest,how long does it take the vehicle to reach this speed?