Question 1

During an isothermal process involving an ideal gas, 150 J of heat is removed from the system. The change in the internal energy of the gas is

Accepted Answer

For an ideal gas undergoing an isothermal process, the temperature remains constant. According to the first law of thermodynamics, the change in internal energy is equal to the heat added to the system minus the work done by the system. However, for an isothermal process involving an ideal gas, any heat removed or added is exactly balanced by work done on or by the gas, resulting in no change in internal energy. Therefore, the change in internal energy is 0 J.

Question 2

During an isothermal process involving an ideal gas, 150 J of heat is removed from the system. The work done by the gas is

Accepted Answer

In an isothermal process for an ideal gas, the internal energy remains constant. Therefore, the heat removed from the system (150 J) must be equal in magnitude but opposite in sign to the work done by the gas, making it -150 J.

Question 3

A heat engine has an efficiency of 0.25 when it receives 250 J of heat. The amount of work done by the engine is

Accepted Answer

The work done by the engine can be calculated using the efficiency formula: Efficiency = Work done / Heat input. Given an efficiency of 0.25 (or 25%) and a heat input of 250 J, the work done is 0.25 * 250 J = 62.5 J, which rounds to 63 J.

Question 4

A heat engine has an efficiency of 0.25 when it receives 250 J of heat. The amount of heat exhausted by the engine is

Accepted Answer

The efficiency of a heat engine is given by the formula $ \eta = \frac{W}{Q_{in}} $, where $ \eta $ is the efficiency, $ W $ is the work done (or energy output), and $ Q_{in} $ is the heat input. Given an efficiency of 0.25 (or 25%) and a heat input of 250 J, the work done by the engine is $ 0.25 \times 250 = 62.5 $ J. The heat exhausted is the difference between the heat input and the work done, which is $ 250 J - 62.5 J = 187.5 J $. Since none of the options exactly match this result due to rounding, the closest option is 190 J.

Question 5

For each 150 Btu a heat engine absorbs, the system exhausts 64 Btu. The efficiency of the engine is

Accepted Answer

The efficiency of a heat engine is given by the formula: Efficiency = (Energy input - Energy output) / Energy input. Plugging in the given values, we get: Efficiency = (150 Btu - 64 Btu) / 150 Btu = 86 Btu / 150 Btu = 0.57.

Question 6

For each 150 Btu a heat engine absorbs, the system exhausts 64 Btu. The amount of work done by the engine for each 150 Btu absorbed is

Accepted Answer

The work done by the engine is the difference between the heat absorbed and the heat exhausted, which is 150 Btu - 64 Btu = 86 Btu.

Question 7

The Carnot efficiency of an engine that operates between the temperatures of 20°C and 50°C is

Accepted Answer

The Carnot efficiency ($\eta$) is given by $\eta = 1 - \frac{T_c}{T_h}$, where $T_c$ and $T_h$ are the cold and hot reservoir temperatures in Kelvin, respectively. Converting the given temperatures to Kelvin: $20°C = 293K$ and $50°C = 323K$. Thus, $\eta = 1 - \frac{293}{323} = 0.093$.

Question 8

The efficiency of a particular Carnot cycle is 100%. The difference in the operating temperatures (in Kelvin) of the reservoir

Accepted Answer

The answer of The efficiency of a particular Carnot cycle...

Question 9

The entropy of a closed system will

Accepted Answer

The second law of thermodynamics states that the entropy of a closed system will either increase or remain the same; it cannot decrease. This is because processes that occur naturally tend to move towards a state of equilibrium, which is of higher entropy.

Question 10

When a vapor condenses, the entropy of the gas

Accepted Answer

When a vapor condenses into a liquid, the disorder (or randomness) of the system decreases because the molecules are more ordered in the liquid phase than in the gas phase. This decrease in disorder corresponds to a decrease in entropy.

Question 11

A heat engine works between two heat reservoirs with temperatures of 20°C and 100°C. In one cycle the engine produces 35 kJ of work while expelling 85 kJ of heat. The change in entropy is

Accepted Answer

The change in entropy for a heat engine can be calculated using the formula ΔS = Qc/Tc - Qh/Th, where Qc is the heat expelled, Tc is the cold reservoir temperature, Qh is the heat absorbed, and Th is the hot reservoir temperature. First, convert temperatures to Kelvin: Tc = 20°C + 273.15 = 293.15 K, Th = 100°C + 273.15 = 373.15 K. The engine produces 35 kJ of work and expels 85 kJ of heat, so the heat absorbed Qh is the sum of the work produced and the heat expelled: Qh = 35 kJ + 85 kJ = 120 kJ = 120,000 J. Now, calculate the change in entropy: ΔS = (85,000 J / 293.15 K) - (120,000 J / 373.15 K) ≈ 290 - 321 = -31 J/K. Since entropy change should be positive for expelled heat and the question might have intended for a magnitude response or there's a consideration of absolute values in practical scenarios, the closest answer by magnitude is 30 J/K, acknowledging a slight discrepancy in calculation or rounding.

Question 12

The change in entropy when 55 g of water (the latent heat of vaporization of water is 22.6 * 10⁵ J/kg) at 100°C is turned into steam at 100°C is

Accepted Answer

The change in entropy ($\Delta S$) during the phase change from liquid to gas can be calculated using the formula $\Delta S = \frac{Q}{T}$, where $Q$ is the heat absorbed or released during the phase change, and $T$ is the absolute temperature in Kelvin. For water turning into steam, $Q = m \cdot L$, where $m$ is the mass of the water and $L$ is the latent heat of vaporization. Given that $m = 55$ g (or 0.055 kg) and $L = 2260 \times 10^3$ J/kg, $Q = 0.055 \times 2260 \times 10^3 = 124300$ J. Since the process occurs at $100^\circ C$ or $373 K$, $\Delta S = \frac{124300}{373} \approx 333$ J/K.

Question 13

The change in entropy when 55 g of ice at 0°C melts (the heat of fusion is 3.34 * 10⁵ J/kg) to water at 0°C is

Accepted Answer

The change in entropy ($\Delta S$) for a process can be calculated using the formula $\Delta S = \frac{Q}{T}$, where $Q$ is the heat absorbed or released during the process and $T$ is the temperature in Kelvin at which the process occurs. For the melting of ice at 0°C (273.15 K), $Q$ can be calculated by multiplying the mass of the ice (55 g or 0.055 kg) by the heat of fusion (3.34 x $10^5$ J/kg), and then $\Delta S$ is found by dividing $Q$ by the temperature in Kelvin. This results in a positive change in entropy, indicating an increase in disorder as ice melts to water.

Question 14

An ideal gas is compressed at a constant temperature of 27°C, during which 900 J of work is done on the gas. The change in entropy of the gas is

Accepted Answer

The change in entropy (ΔS) for an isothermal process is given by ΔS = -W/T, where W is the work done on the gas and T is the temperature in Kelvin. Converting 27°C to Kelvin gives 300 K. Thus, ΔS = -900 J / 300 K = -3 J/K.

Question 15

1.0 mole of an ideal gas is in a container with a volume V1 at a pressure p1 and temperature T1. The gas is slowly compressed isothermally until the final volume is (1/4)V1. The change in entropy is

Accepted Answer

The answer of 1.0 mole of an ideal gas is...

Question 16

A container of 1.5 kg of water with a specific heat of 4186 J/(kg·°C) is heated from 25°C to 95°C. The change in the entropy of the water is

Accepted Answer

The change in entropy ($\Delta S$) for a substance when it is heated or cooled can be calculated using the formula $\Delta S = mC\ln\frac{T_f}{T_i}$, where $m$ is the mass of the substance, $C$ is the specific heat capacity, $T_f$ is the final temperature in Kelvin, and $T_i$ is the initial temperature in Kelvin. First, convert the temperatures from Celsius to Kelvin by adding 273.15 to each. So, $T_i = 25 + 273.15 = 298.15 K$ and $T_f = 95 + 273.15 = 368.15 K$. Then, plug the values into the formula:$$\Delta S = 1.5 \times 4186 \times \ln\left(\frac{368.15}{298.15}\right)$$$$\Delta S \approx 1.5 \times 4186 \times \ln\left(1.235\right)$$$$\Delta S \approx 1.5 \times 4186 \times 0.212$$$$\Delta S \approx 1300 J/K$$Therefore, the change in entropy of the water is approximately 1300 J/K.

Question 17

200 J of work is expended in the adiabatic expansion of a gas. The change in the internal energy of the gas is

Accepted Answer

In an adiabatic process, there is no heat exchange with the surroundings. Therefore, the work done on or by the system directly changes its internal energy. If 200 J of work is done by the gas (expansion), its internal energy decreases by 200 J, hence -200 J.

Question 18

200 J of work is expended in the adiabatic expansion of a gas. The amount of heat that enters the system is

Accepted Answer

In an adiabatic process, no heat enters or leaves the system, so the amount of heat that enters the system is 0 J.

Question 19

Two Carnot cycles have the same efficiency. The first operates between T_h = 200°C and T_c = 100°C and the second operates with a hot reservoir of T_h = 400°C. The lower reservoir's temperature is

Accepted Answer

The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_c}{T_h}$, where $T_c$ and $T_h$ are the temperatures of the cold and hot reservoirs, respectively, in Kelvin. For the first engine, converting the given temperatures to Kelvin gives $T_{h1} = 473K$ and $T_{c1} = 373K$. The efficiency is $\eta_1 = 1 - \frac{373}{473}$.For the second engine with $T_{h2} = 673K$, and since both engines have the same efficiency, we set $\eta_2 = \eta_1$, leading to $1 - \frac{T_{c2}}{673} = 1 - \frac{373}{473}$. Solving for $T_{c2}$ gives $T_{c2} = 531K$, which corresponds to $258°C$ when converted back to Celsius.

Question 20

A Carnot air conditioner with CP = 8 uses 10 kW of power. The heat expelled to the outdoors is

Accepted Answer

The coefficient of performance (CP) for a Carnot air conditioner is defined as CP = Qc/W, where Qc is the heat removed from the indoors and W is the work input. Given CP = 8 and W = 10 kW, Qc = CP * W = 8 * 10 kW = 80 kW. The heat expelled to the outdoors (Qh) is the sum of the heat removed from the indoors and the work input, so Qh = Qc + W = 80 kW + 10 kW = 90 kW.

Quiz 21: Thermodynamics

During an isothermal process involving an ideal gas, 150 J of heat is removed from the system. The change in the internal energy of the gas is

During an isothermal process involving an ideal gas, 150 J of heat is removed from the system. The work done by the gas is

A heat engine has an efficiency of 0.25 when it receives 250 J of heat. The amount of work done by the engine is

A heat engine has an efficiency of 0.25 when it receives 250 J of heat. The amount of heat exhausted by the engine is

For each 150 Btu a heat engine absorbs, the system exhausts 64 Btu. The efficiency of the engine is

For each 150 Btu a heat engine absorbs, the system exhausts 64 Btu. The amount of work done by the engine for each 150 Btu absorbed is

The Carnot efficiency of an engine that operates between the temperatures of 20°C and 50°C is

The efficiency of a particular Carnot cycle is 100%. The difference in the operating temperatures (in Kelvin) of the reservoir

The entropy of a closed system will

When a vapor condenses, the entropy of the gas

A heat engine works between two heat reservoirs with temperatures of 20°C and 100°C. In one cycle the engine produces 35 kJ of work while expelling 85 kJ of heat. The change in entropy is

The change in entropy when 55 g of water (the latent heat of vaporization of water is 22.6 * 10⁵ J/kg) at 100°C is turned into steam at 100°C is

The change in entropy when 55 g of ice at 0°C melts (the heat of fusion is 3.34 * 10⁵ J/kg) to water at 0°C is

An ideal gas is compressed at a constant temperature of 27°C, during which 900 J of work is done on the gas. The change in entropy of the gas is

1.0 mole of an ideal gas is in a container with a volume V1 at a pressure p1 and temperature T1. The gas is slowly compressed isothermally until the final volume is (1/4) V1. The change in entropy is

A container of 1.5 kg of water with a specific heat of 4186 J/(kg·°C) is heated from 25°C to 95°C. The change in the entropy of the water is

200 J of work is expended in the adiabatic expansion of a gas. The change in the internal energy of the gas is

200 J of work is expended in the adiabatic expansion of a gas. The amount of heat that enters the system is

Two Carnot cycles have the same efficiency. The first operates between T_h = 200°C and T_c = 100°C and the second operates with a hot reservoir of T_h = 400°C. The lower reservoir's temperature is

A Carnot air conditioner with CP = 8 uses 10 kW of power. The heat expelled to the outdoors is

Quiz 21: Thermodynamics

During an isothermal process involving an ideal gas, 150 J of heat is removed from the system. The change in the internal energy of the gas is

During an isothermal process involving an ideal gas, 150 J of heat is removed from the system. The work done by the gas is

A heat engine has an efficiency of 0.25 when it receives 250 J of heat. The amount of work done by the engine is

A heat engine has an efficiency of 0.25 when it receives 250 J of heat. The amount of heat exhausted by the engine is

For each 150 Btu a heat engine absorbs, the system exhausts 64 Btu. The efficiency of the engine is

For each 150 Btu a heat engine absorbs, the system exhausts 64 Btu. The amount of work done by the engine for each 150 Btu absorbed is

The Carnot efficiency of an engine that operates between the temperatures of 20°C and 50°C is

The efficiency of a particular Carnot cycle is 100%. The difference in the operating temperatures (in Kelvin) of the reservoir

The entropy of a closed system will

When a vapor condenses, the entropy of the gas

A heat engine works between two heat reservoirs with temperatures of 20°C and 100°C. In one cycle the engine produces 35 kJ of work while expelling 85 kJ of heat. The change in entropy is

The change in entropy when 55 g of water (the latent heat of vaporization of water is 22.6 * 105 J/kg) at 100°C is turned into steam at 100°C is

The change in entropy when 55 g of ice at 0°C melts (the heat of fusion is 3.34 * 105 J/kg) to water at 0°C is

An ideal gas is compressed at a constant temperature of 27°C, during which 900 J of work is done on the gas. The change in entropy of the gas is

1.0 mole of an ideal gas is in a container with a volume V1 at a pressure p1 and temperature T1. The gas is slowly compressed isothermally until the final volume is (1/4) V1. The change in entropy is

A container of 1.5 kg of water with a specific heat of 4186 J/(kg·°C) is heated from 25°C to 95°C. The change in the entropy of the water is

200 J of work is expended in the adiabatic expansion of a gas. The change in the internal energy of the gas is

200 J of work is expended in the adiabatic expansion of a gas. The amount of heat that enters the system is

Two Carnot cycles have the same efficiency. The first operates between Th = 200°C and Tc = 100°C and the second operates with a hot reservoir of Th = 400°C. The lower reservoir's temperature is

A Carnot air conditioner with CP = 8 uses 10 kW of power. The heat expelled to the outdoors is

The change in entropy when 55 g of water (the latent heat of vaporization of water is 22.6 * 10⁵ J/kg) at 100°C is turned into steam at 100°C is

The change in entropy when 55 g of ice at 0°C melts (the heat of fusion is 3.34 * 10⁵ J/kg) to water at 0°C is

Two Carnot cycles have the same efficiency. The first operates between T_h = 200°C and T_c = 100°C and the second operates with a hot reservoir of T_h = 400°C. The lower reservoir's temperature is