Question 1

A Carnot refrigerator has a coefficient of performance of 2.5. The refrigerator consumes 50 W of power. How much heat is removed from the interior of the refrigerator in 1 hour?

Accepted Answer

The coefficient of performance (COP) for a refrigerator is defined as the ratio of the heat removed from the refrigerator (Q_c) to the work input (W). Given COP = 2.5 and W = 50 W, we can calculate Q_c as follows: COP = Q_c / W, so Q_c = COP * W = 2.5 * 50 W = 125 W. To find the heat removed in 1 hour, convert power (in watts) to energy (in joules) using the conversion 1 W = 1 J/s. Thus, 125 J/s * 3600 s = 450,000 J or 450 kJ.

Question 2

A Carnot refrigerator takes heat from water at 0°C and rejects heat to a room at 12°C. Suppose that 92.0 grams of water at 0°C are converted to ice at 0°C by the refrigerator. Calculate the mechanical energy that must be supplied to the refrigerator. The heat of fusion of water is 3.34 × 10⁵ J/kg.

Accepted Answer

The answer of A Carnot refrigerator takes heat from water...

Question 3

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C exhaust temperature = 430°C 7.0 × 10⁹ cal of fuel produces 1.4 × 10¹⁰ J of work in one hour, where 1 cal = 4.19 J (a) What is the actual efficiency of this engine? (b) What is the Carnot efficiency of the engine? (c) What is the power output of this engine?

Accepted Answer

The answer of One of the most efficient engines built...

Question 4

A certain Carnot heat pump transfers energy from the outside of a house, where the temperature is -15°C, to the inside of a room where the temperature is 21°C. If this heat pump runs off of electricity, what is the minimum rate at which it uses electrical energy to deliver 150 W to the inside room?

Accepted Answer

The answer of A certain Carnot heat pump transfers energy...

Question 5

An ideal reversible refrigerator keeps its inside compartment at 9.0°C. What is the high temperature, Th, needed to give this refrigerator a coefficient of performance of 3.7?

Accepted Answer

The coefficient of performance of a refrigerator is given by:$$COP = \frac{Th}{Th - Tc}$$where Th is the high temperature and Tc is the low temperature.In this case, the COP is 3.7 and the low temperature is 9.0°C. Solving for Th, we get:$$Th = Tc * COP + Tc$$$$Th = 9.0°C * 3.7 + 9.0°C$$$$Th = 42.3°C$$The closest answer choice is A) 85°C.

Question 6

A Carnot engine operating between a reservoir of liquid mercury at its melting point (233 K) and a colder reservoir extracts 10.0 J of heat from the mercury and does 8.0 J of work during each cycle. What is the temperature of the colder reservoir?

Accepted Answer

The efficiency of a Carnot engine is given by:$$e = 1 - \frac{T_c}{T_h}$$where $T_c$ is the temperature of the colder reservoir and $T_h$ is the temperature of the hotter reservoir.We are given that the efficiency of the engine is 80%, so we can write:$$0.8 = 1 - \frac{T_c}{233 K}$$Solving for $T_c$, we get:$$T_c = 233 K - 0.8 * 233 K = 46.6 K$$Therefore, the temperature of the colder reservoir is 46.6 K, which is closest to choice A, 47 K.

Question 7

A perfect Carnot engine operates between the temperatures of 300K and 700K, drawing 60 kJ of heat from the 700K reservoir in each cycle. How much heat is dumped into the 300K reservoir in each cycle?

Accepted Answer

The answer of A perfect Carnot engine operates between the...

Question 8

A Carnot engine is operated as a heat pump to heat a room in the winter. The heat pump delivers heat to the room at the rate of 34 kJ per second and maintains the room at a temperature of 293 K when the outside temperature is 229 K. The power requirement for the heat pump under these operating conditions is closest to

Accepted Answer

The power requirement for the heat pump is given by the following equation:$$P = \frac{Q}{COP}$$where:* P is the power requirement in watts* Q is the heat delivered to the room in joules per second* COP is the coefficient of performance of the heat pumpThe COP of a Carnot heat pump is given by the following equation:$$COP = \frac{T_H}{T_H - T_C}$$where:* T_H is the temperature of the hot reservoir in kelvins* T_C is the temperature of the cold reservoir in kelvinsIn this problem, we are given the following information:* Q = 34 kJ/s* T_H = 293 K* T_C = 229 KSubstituting these values into the equations above, we get:$$COP = \frac{293 K}{293 K - 229 K} = 2.23$$$$P = \frac{34 kJ/s}{2.23} = 15.25 kW$$Therefore, the power requirement for the heat pump is closest to 15,250 W, which is choice A.

Question 9

You want to design an ideal Carnot heat engine that wastes only 35.0% of the heat that goes into it. The lowest cold-reservoir temperature available to you is +15.0°C. If 150.0 J of work is done per cycle, the heat input per cycle is closest to

Accepted Answer

The efficiency of a Carnot engine is given by η = 1 - (Tc/Th), where Tc and Th are the absolute temperatures of the cold and hot reservoirs, respectively. If 35% of the heat is wasted, the efficiency is 65% or 0.65. The work done (W) is 150 J, and efficiency is also W/Qh, where Qh is the heat input. Solving 0.65 = 150/Qh gives Qh ≈ 231 J.

Question 10

A refrigerator has a coefficient of performance of 1.15, and it extracts 7.95 J of heat from the cold reservoir during each cycle.
(a) How much work is done on the gas in each cycle?
(b) How much heat is exhausted into the hot reservoir in each cycle?

Accepted Answer

The answer of A refrigerator has a coefficient of performance...

Question 11

An ideal Carnot engine operates between reservoirs having temperatures of 125°C and -20°C. Each cycle the heat expelled by this engine is used to melt 30.0 g of ice at 0.00°C. The heat of fusion of water is 3.34 × 10⁵ J/kg and the heat of vaporization of water is 2.25 × 10⁶ J/kg. (a) How much work does this engine do each cycle? (b) How much heat per cycle does this engine absorb at the hot reservoir?

Accepted Answer

The answer of An ideal Carnot engine operates between reservoirs...

Question 12

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 13

What is the maximum theoretical efficiency possible for a heat engine operating between a reservoir in which ice and water coexist, and a reservoir in which water and steam coexist? The pressure is constant at 1.0 atmosphere for both reservoirs.

Accepted Answer

The maximum theoretical efficiency of a heat engine is given by the Carnot efficiency, which is calculated using the temperatures of the hot and cold reservoirs (Th and Tc, respectively) in Kelvin. The temperature at which ice and water coexist (the melting point of ice) is 0°C or 273K, and the temperature at which water and steam coexist (the boiling point of water) is 100°C or 373K under 1.0 atmosphere pressure. The Carnot efficiency is given by 1 - (Tc/Th), so the efficiency is 1 - (273/373) = 0.268, or 26.8%, which rounds to approximately 27%.

Question 14

The temperature inside a Carnot refrigerator placed in a kitchen at 22.0°C is 2.0°C. The heat extracted from the refrigerator is 89 MJ/h. What power is needed to operate this refrigerator?

Accepted Answer

We can use the Carnot efficiency formula to find the heat rejected by the refrigerator:
ε = 1 - T_cold/T_hot
where ε is the Carnot efficiency, T_cold is the temperature inside the refrigerator, and T_hot is the temperature outside the refrigerator (in the kitchen). Plugging in the given values, we get:
ε = 1 - 2.0°C / 22.0°C = 0.9091
This means that the heat rejected by the refrigerator is 90.91% of the heat extracted, or:
Q_rejected = 0.9091 x 89 MJ/h = 81 MJ/h
The power needed to operate the refrigerator is then the net heat extracted divided by the Carnot efficiency:
P = (89 MJ/h - 81 MJ/h) / 0.9091 = 8 MJ/h / 0.9091 = 8.8 kW/h / 0.9091 = 1.83 kW
Rounding to two significant figures, we get:
Answer: B (1.8 kW)

Question 15

A Carnot engine operates between a high temperature reservoir at 435 K and a river with water at 280 K. If it absorbs 3700 J of heat each cycle, how much work per cycle does it perform?

Accepted Answer

The efficiency of a Carnot engine is given by:$$e=1-\frac{T_c}{T_h}$$where $T_c$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir.In this case, $T_c=280$ K and $T_h=435$ K, so:$$e=1-\frac{280}{435}=0.356$$The work done per cycle is given by:$$W=eQ_h$$where $Q_h$ is the heat absorbed per cycle.In this case, $Q_h=3700$ J, so:$$W=0.356	imes3700	ext{ J}=1318	ext{ J}$$Therefore, the correct answer is A.

Question 16

A coal-fired plant generates 600 MW of electric power. The plant uses 4.8 × 10⁶ kg of coal each day. The heat produced by the combustion of coal is 3.3 × 10⁷ J/kg. The steam that drives the turbines is at a temperature of 300°C, and the exhaust water is at 37°C. (a) What is the overall efficiency of the plant for generating electric power? (b) What is the maximum efficiency that this plant could possibly have using the same temperature extremes that it presently uses? (c) How much thermal energy is exhausted each day by this plant?

Accepted Answer

The answer of A coal-fired plant generates 600 MW of...

Question 17

A Carnot cycle engine operates between a low temperature reservoir at 20°C and a high temperature reservoir at 800°C. If the engine is required to output 20.0 kJ of work per cycle, how much heat must the high temperature reservoir transfer to the engine during each cycle?

Accepted Answer

The efficiency of a Carnot cycle engine is given by:$$e = 1 - \frac{T_L}{T_H}$$where $T_L$ is the temperature of the low temperature reservoir and $T_H$ is the temperature of the high temperature reservoir.In this case, $T_L = 20°C + 273.15 = 293.15 K$ and $T_H = 800°C + 273.15 = 1073.15 K$. Therefore, the efficiency is:$$e = 1 - \frac{293.15 K}{1073.15 K} = 0.729$$The work output per cycle is given by:$$W = eQ_H$$where $Q_H$ is the heat transferred to the engine from the high temperature reservoir.In this case, $W = 20.0 kJ$. Therefore, the heat transferred to the engine from the high temperature reservoir is:$$Q_H = \frac{W}{e} = \frac{20.0 kJ}{0.729} = 27.5 kJ$$

Question 18

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 19

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 20

A Carnot engine operates between reservoirs at 550K and 300K, discarding 1500 J of heat in each cycle.
(a) What is the engine's efficiency?
(b) How much heat is supplied to the engine by the hot reservoir in each cycle?

Accepted Answer

The answer of A Carnot engine operates between reservoirs at...

Quiz 19: The Second Law of Thermodynamics