Question 1

An ideal, reversible heat pump is taking heat from the outside air at -10.0°C and discharging it into the house at 18.0°C. What is the coefficient of performance of this heat pump?

Accepted Answer

The answer of An ideal, reversible heat pump is taking...

Question 2

An irreversible engine operating between the temperatures of 550 K and 300 K extracts 1200 J of heat from the hot reservoir and produces 450 J of work. How much entropy is created in the process?

Accepted Answer

Entropy created in the process is given by:ΔS = Qh/Th - Qc/Tc= 1200 J / 550 K - 450 J / 300 K= 0.32 J/K

Question 3

An air conditioner with a coefficient of performance of 3.5 uses 30 kW of power. How much power is it discharging to the outdoors?

Accepted Answer

The answer of An air conditioner with a coefficient of...

Question 4

On a cold winter day, the outside temperature is -20°C and the inside temperature is maintained at 20°C. There is a net heat flow to the outside through the walls, roof, etc., of 25 kW. What is the rate of increase of the entropy of the universe as a result of this?

Accepted Answer

The rate of increase of entropy of the universe is calculated using the formula: ΔS = Q/T. The heat flow is 25 kW, and the temperatures are 20°C (293 K) inside and -20°C (253 K) outside. The entropy increase is due to the heat flow to the colder reservoir: ΔS = 25000 W / 253 K - 25000 W / 293 K ≈ 13 W/K.

Question 5

A Carnot engine has an efficiency of 83.0% and performs 4500 J of work every cycle. How much energy is discharged to the lower temperature reservoir every cycle?

Accepted Answer

The answer of A Carnot engine has an efficiency of...

Question 6

What is the change in entropy when 15.0 g of water at 100°C are turned into steam at 100°C? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg.

Accepted Answer

The change in entropy (ΔS) is calculated using the formula ΔS = Q/T, where Q is the heat absorbed or released and T is the temperature in Kelvin. The mass of water is 0.015 kg, and the latent heat of vaporization is 22.6 × 10^5 J/kg. Therefore, Q = 0.015 kg × 22.6 × 10^5 J/kg = 33900 J. The temperature in Kelvin is 373 K. Thus, ΔS = 33900 J / 373 K = 90.8 J/K.

Question 7

The compressor in a certain Carnot refrigerator performs 480 J of work to remove 150 J of heat from the interior of the refrigerator. How much heat must the coils behind the refrigerator discharge into the kitchen?

Accepted Answer

The Carnot refrigerator operates on the principle of conservation of energy. The work done by the compressor (480 J) is used to remove heat from the interior (150 J) and the rest is discharged as heat into the surroundings. Therefore, the heat discharged is the sum of the work done and the heat removed from the interior: 480 J + 150 J = 630 J.

Question 8

What is the maximum efficiency of an engine operating between 500 K and 300 K?

Accepted Answer

The maximum efficiency of an engine operating between 500 K and 300 K can be calculated using the Carnot efficiency formula, which is given by:
Efficiency = 1 - (T_low / T_high)
where T_low is the temperature of the cold reservoir (in Kelvin) and T_high is the temperature of the hot reservoir (in Kelvin).
Substituting the given values, we get:
Efficiency = 1 - (300 / 500) = 1 - 0.6 = 0.4 or 40%
Therefore, the best choice is C, which corresponds to an efficiency of 40%.

Question 9

A Carnot air conditioner has a coefficient of performance of 17.0 and removes 72.0 MJ of heat from the interior of a house every hour. How much power does it consume?

Accepted Answer

The coefficient of performance (COP) for a Carnot air conditioner is defined as the ratio of the heat removed (Q_c) from the cold reservoir to the work input (W), i.e., COP = Q_c/W. Given COP = 17.0 and Q_c = 72.0 MJ/h = 72000 kJ/h, we can rearrange the formula to solve for W: W = Q_c/COP. Converting 72000 kJ/h to watts (since power is in watts and 1 W = 1 J/s), we first convert hours to seconds (72000 kJ/h * 1000 J/kJ) / (3600 s) = 20000 W = 20 kW. Then, W = 20 kW / 17.0 = 1176.47 W, which rounds to approximately 1180 W.

Question 10

A Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does it need to remove 2000 J of heat from the interior of the house?

Accepted Answer

The answer of A Carnot air conditioner operates between an...

Question 11

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 12

A heat engine operating at maximum efficiency has an efficiency of 25%. The temperature of the cold reservoir is 300 K. What is the temperature of the hot reservoir?

Accepted Answer

The maximum efficiency of a heat engine is given by the equation:

Efficiency = 1 - (Tc/Th)

Where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.

Plugging in the given values:

0.25 = 1 - (300/Th)

0.75 = 300/Th

Th = 400 K

Therefore, the temperature of the hot reservoir is 400 K, which corresponds to choice C).

Question 13

A heat engine operating at maximum efficiency has an efficiency of 35.0%. The temperature of the hot reservoir is 700 K. What is the temperature of the cold reservoir?

Accepted Answer

The answer of A heat engine operating at maximum efficiency...

Question 14

On a cold winter day, the outside temperature is -20°C and the inside temperature is maintained at 20°C. There is a net heat flow to the outside through the walls, roof, etc., of 25 kW. At what rate is the air outside the house gaining entropy as a result of this process?

Accepted Answer

The rate of entropy gain is given by the heat flow divided by the temperature difference. In this case, the heat flow is 25 kW and the temperature difference is 40°C, so the rate of entropy gain is 25 kW / 40°C = 0.625 kW/K = 625 W/K.

Question 15

The ocean thermal energy conversion project uses the surface water near tropical islands with a temperature of 20.0°C as the hot temperature reservoir, and the water at some depth, with a temperature of 5.0°C, as the cold temperature reservoir for a heat engine. What is the maximum efficiency of an engine running between those two temperatures?

Accepted Answer

The answer of The ocean thermal energy conversion project uses...

Question 16

A certain Carnot engine has T_h = T_c + 35K, and an efficiency of 20.0%. What is T_h?

Accepted Answer

We can start by using the equation for the efficiency of a Carnot engine:

efficiency = 1 - (T2/T1)

where T2 is the temperature of the cold reservoir and T1 is the temperature of the hot reservoir.

We are given that the efficiency of this Carnot engine is 20.0%, or 0.20. Therefore, we can write:

0.20 = 1 - (T2/T1)

Solving for T2/T1, we get:

T2/T1 = 0.80

We are also given that:

TS1U1B1hS1U1B0 = TS1U1B1cS1U1B0 + 35K

This tells us that the temperature of the hot reservoir (TS1U1B1h) is 35 K higher than the temperature of the cold reservoir (TS1U1B1c). We can use these values to set up equations for T1 and T2:

T1 = TS1U1B1hS1U1B0
T2 = TS1U1B1cS1U1B0

Using T2/T1 = 0.80, we can solve for T2:

T2 = (0.80)T1

Substituting in the values for T1 and T2 in terms of the given temperatures, we get:

TS1U1B1cS1U1B0 = T2 = (0.80)T1 = (0.80)(TS1U1B1hS1U1B0)
TS1U1B1hS1U1B0 = TS1U1B1cS1U1B0 + 35K

Substituting the first equation into the second equation, we get:

TS1U1B1hS1U1B0 = (0.80)(TS1U1B1hS1U1B0) + 35K

Solving for TS1U1B1hS1U1B0, we get:

TS1U1B1hS1U1B0 = 175K

Therefore, the answer is C.

Question 17

A 40.0-L container is divided into two equal parts by a rubber membrane. One half of the container has 1.80 moles of an ideal monatomic gas at 270 K, and the other half is a vacuum. The container does not allow any gas to escape but is free to exchange heat with the surroundings. The membrane breaks, the gas expands isothermally and eventually the gas reaches a new equilibrium condition occupying the entire volume. What is the change in entropy of the gas?

Accepted Answer

The change in entropy is given by:$$\Delta S = nR\ln(\frac{V_f}{V_i})$$where:* n is the number of moles of gas* R is the ideal gas constant* V_f is the final volume* V_i is the initial volumeIn this case, n = 1.80 moles, R = 8.31 J/mol*K, V_f = 40.0 L, and V_i = 20.0 L. Substituting these values into the equation, we get:$$\Delta S = 1.80 mol * 8.31 J/mol*K * \ln(\frac{40.0 L}{20.0 L}) = 10.4 J/K$$

Quiz 18: The Laws of Thermodynamics