Thermodynamics Quiz: Cop For Refrigerators And Heat Pumps

20 questions · exam conditions

0:00

Cop For Refrigerators And Heat PumpsQuestion 1 of 20

A Carnot refrigerator keeps contents at -10°C while the kitchen is 20°C. Its COP is closest to

8.8

9.8

0.10

-0.33

← Back to quizzes

Thermodynamics Quiz

Thermodynamics Quiz: Cop For Refrigerators And Heat Pumps

Practice Cop For Refrigerators And Heat Pumps in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Cop For Refrigerators And Heat Pumps, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A Carnot refrigerator keeps contents at -10°C while the kitchen is 20°C. Its COP is closest to

8.8 (correct answer)
9.8
0.10
-0.33

Explanation: Convert to kelvin: the cold reservoir is 263 K and the warm reservoir is 293 K. A Carnot refrigerator's COP is the cold temperature divided by the temperature difference, so COP = 263 / (293 - 263) = 263 / 30 = 8.8. The nearby wrong value 9.8 comes from using the warm kitchen temperature in the numerator, but the refrigerator COP uses the cold side.

Question 2

A Carnot heat pump keeps a house at 20°C with outside air at -5°C. It delivers 15 kW. Power input is closest to

0.60 kW
1.40 kW
1.02 kW
1.28 kW (correct answer)

Explanation: A Carnot heat pump's COP is hot temperature over temperature difference: 293 / 25 = 11.72. Delivering 15 kW to the house requires 15 / 11.72 = 1.28 kW. Using 268 / 25 = 10.72 gives 1.40 kW, but that treats 15 kW as heat removed from outdoors rather than delivered indoors.

Question 3

A refrigerator removes 2 kW from its cold space while drawing 0.5 kW. As a heat pump under the same rates, its COP is

4.0
5.0 (correct answer)
1.25
0.20

Explanation: A heat pump's COP is heat delivered to the warm space divided by work input. The heat delivered is the 2 kW removed from the cold space plus the 0.5 kW of work added, so 2.5 / 0.5 = 5.0. The tempting 4.0 is the refrigerator COP, which uses only the 2 kW removed.

Question 4

A reversible engine between 250 K and 300 K outputs 1 kW net. Run in reverse as a refrigerator, its cooling rate is

0.83 kW
5.0 kW (correct answer)
6.0 kW
1.0 kW

Explanation: A reversible engine's efficiency is 1 - 250/300 = 1/6, so 1 kW output means it absorbs 6 kW from the hot reservoir and rejects 5 kW to the cold one. Reversed as a refrigerator, the cooling rate is that same 5 kW. The tempting 0.83 kW mistake comes from using efficiency as W divided by the cold-side heat instead of the hot-side heat.

Question 5

A refrigerator removes 300 kJ from its contents and rejects 400 kJ to the room. Its COP is

0.75
1.33
3.0 (correct answer)
4.0

Explanation: A refrigerator's COP is heat removed from the cold contents divided by the work input. The work input is the extra heat rejected beyond what was removed: 400 - 300 = 100 kJ. So COP = 300 / 100 = 3.0. Don't divide 400 by 300; that ignores the work input and gives 1.33.

Question 6

A heat pump operating between two thermal reservoirs has a COP of $4.2$ . If it delivers $2100 \text{ kJ}$ to the hot reservoir, what is the work input required?

500 kJ (correct answer)
600 kJ
400 kJ
300 kJ
700 kJ

Explanation: When you encounter heat pump problems, focus on the relationship between coefficient of performance (COP), work input, and heat delivery. A heat pump's COP is defined as the ratio of heat delivered to the hot reservoir (

Q_H

) to the work input (

W

\text{COP} = \frac{Q_H}{W}

. Given that the COP is 4.2 and the heat delivered is 2100 kJ, you can solve for work input by rearranging the COP equation:

W = \frac{Q_H}{\text{COP}} = \frac{2100 \text{ kJ}}{4.2} = 500 \text{ kJ}

. Looking at the incorrect answers: Answer B (600 kJ) likely comes from incorrectly using the refrigerator COP formula or making an arithmetic error. Answer C (400 kJ) might result from confusing the heat pump COP with other thermodynamic ratios or computational mistakes. Answer D (300 kJ) could stem from misapplying the relationship between heat absorbed from the cold reservoir and work, or from incorrectly manipulating the COP formula. The correct answer is A (500 kJ), which directly follows from the fundamental heat pump COP definition. Remember that heat pump COP is always the ratio of useful heat output to work input. This differs from refrigerator COP, which uses heat removed from the cold reservoir. Also, heat pump COP values are typically greater than 1 (often 3-6 for real systems), making them more efficient than electric resistance heating. Always double-check your COP formula based on whether you're dealing with a heat pump or refrigerator.

Question 7

A heat pump has a COP of $5.5$ when operating between the same two thermal reservoirs. What would be the COP if this system operates as a refrigerator?

4.5 (correct answer)
6.5
5.5
0.818
0.182

Explanation: When you encounter heat pump and refrigerator COP problems, remember that these are the same device operating with different objectives - the key is understanding how their COPs relate mathematically. A heat pump and refrigerator operating between the same thermal reservoirs are thermodynamically identical systems with different purposes. The heat pump delivers heat to the hot reservoir, while the refrigerator removes heat from the cold reservoir. Their COPs are related by:

COP_{HP} = COP_{R} + 1

. Given that the heat pump COP is 5.5, we can find the refrigerator COP:

COP_{R} = COP_{HP} - 1 = 5.5 - 1 = 4.5

This confirms answer A is correct. Let's examine why the other options are wrong: B) 6.5 represents adding 1 to the heat pump COP instead of subtracting, which reverses the relationship between the two systems. C) 5.5 assumes both systems have identical COPs, ignoring the fundamental difference in their energy accounting - the heat pump includes both useful heating and refrigeration work in its efficiency measure. D) 0.818 appears to be the reciprocal of 5.5 divided by some factor, which has no physical basis in thermodynamic COP relationships. Remember this essential relationship:

COP_{HP} = COP_{R} + 1

. The heat pump always has a COP exactly 1 unit higher because it delivers the same cooling effect plus the work input as useful heating. Master this simple relationship and you'll quickly solve any heat pump/refrigerator conversion problem.

Question 8

A refrigerator removes $850 \text{ kJ}$ of heat from the cold reservoir and requires $320 \text{ kJ}$ of work input during one complete cycle. What is the coefficient of performance (COP) for this refrigerator?

2.66 (correct answer)
3.66
0.376
0.274
1.66

Explanation: When you encounter refrigerator problems, focus on the coefficient of performance (COP), which measures how efficiently the refrigerator removes heat compared to the work required to operate it. For a refrigerator, the COP is defined as the ratio of heat removed from the cold reservoir (

Q_C

) to the work input (

W

\text{COP} = \frac{Q_C}{W}

. This tells you how many kilojoules of heat are removed for every kilojoule of work invested. Using the given values:

Q_C = 850 \text{ kJ}

and

W = 320 \text{ kJ}

, so

\text{COP} = \frac{850}{320} = 2.66

. This means the refrigerator removes 2.66 kJ of heat for every 1 kJ of work input. Looking at the wrong answers: Choice B (3.66) likely comes from incorrectly adding the work to the heat removed before dividing:

\frac{850 + 320}{320}

. Choice C (0.376) results from flipping the COP formula to

\frac{W}{Q_C} = \frac{320}{850}

, which would give you the reciprocal. Choice D (0.274) comes from using the heat pump COP formula incorrectly:

\frac{W}{Q_C + W} = \frac{320}{1170}

. Remember that refrigerator COP is always

\frac{\text{heat removed}}{\text{work input}}

, and typical values range from 2-5 for household refrigerators. If your calculated COP is less than 1, you've likely made a formula error since that would mean the refrigerator is removing less heat than the work required to run it.

Question 9

A refrigerator has a COP of $3.5$ . If the work input is $150 \text{ kJ}$ , how much heat is rejected to the hot reservoir?

525 kJ
375 kJ
675 kJ (correct answer)
150 kJ
225 kJ

Explanation: Refrigerator problems test your understanding of the coefficient of performance (COP) and energy conservation in thermodynamic cycles. When you see COP questions, remember that refrigerators move heat from a cold space to a hot space using work input. For a refrigerator, COP is defined as the ratio of cooling effect (heat removed from cold reservoir,

Q_C

) to work input:

\text{COP} = \frac{Q_C}{W}

. Given COP = 3.5 and

W = 150 \text{ kJ}

, you can find:

Q_C = \text{COP} \times W = 3.5 \times 150 = 525 \text{ kJ}

The key insight is applying energy conservation. The first law requires that energy input equals energy output:

W + Q_C = Q_H

, where

Q_H

is heat rejected to the hot reservoir. Therefore:

Q_H = 150 + 525 = 675 \text{ kJ}

Answer A (525 kJ) represents

Q_C

, the heat removed from the cold reservoir—this is what you'd get if you stopped after calculating the cooling effect. Answer B (375 kJ) might result from incorrectly subtracting instead of adding:

525 - 150 = 375

. Answer D (150 kJ) is simply the work input, showing confusion about what the question asks for. Remember this pattern: for refrigerator COP problems, first find

Q_C

using the COP definition, then apply energy conservation (

Q_H = W + Q_C

) to find heat rejection. The heat rejected is always larger than both the work input and the cooling effect individually.

Question 10

A refrigerator operates with a COP of $2.8$ . If the same system operates as a heat pump between the same two reservoirs, what would be its COP as a heat pump?

3.8 (correct answer)
2.8
1.8
4.8
0.357

Explanation: When you encounter problems involving refrigerators and heat pumps operating between the same reservoirs, remember that these are the same device running in opposite directions, and their COPs are related by a simple formula. For any system operating between two thermal reservoirs, the relationship between refrigerator COP and heat pump COP is:

\text{COP}_{\text{HP}} = \text{COP}_{\text{R}} + 1

. This comes from the fundamental definitions: a refrigerator's COP is the ratio of heat removed from the cold reservoir to work input (

Q_C/W

), while a heat pump's COP is the ratio of heat delivered to the hot reservoir to work input (

Q_H/W

). Since energy conservation requires

Q_H = Q_C + W

, we get

\text{COP}_{\text{HP}} = Q_H/W = (Q_C + W)/W = Q_C/W + 1 = \text{COP}_{\text{R}} + 1

. Given

\text{COP}_{\text{R}} = 2.8

, the heat pump COP is

2.8 + 1 = 3.8

, making A correct. Choice B (2.8) incorrectly assumes the COPs are identical, missing that heat pumps deliver more total energy than refrigerators extract. Choice C (1.8) mistakenly subtracts 1 instead of adding it. Choice D (4.8) might result from incorrectly adding 2 instead of 1, perhaps confusing this relationship with another thermodynamic formula. Study tip: Always remember "COP heat pump = COP refrigerator + 1" for the same reservoirs. This +1 relationship appears frequently on thermodynamics exams and reflects that heat pumps deliver both the extracted energy and the work energy to the hot reservoir.

Question 11

A refrigerator consumes $180 \text{ kJ}$ of work and has a COP of $3.2$ . How much total heat is transferred to the environment (hot reservoir)?

576 kJ
396 kJ
756 kJ (correct answer)
216 kJ
180 kJ

Explanation: When you encounter refrigerator problems, remember that refrigerators move heat from a cold space to a hot environment, requiring work input. The coefficient of performance (COP) relates the useful cooling effect to the work consumed. For a refrigerator,

\text{COP} = \frac{Q_C}{W}

, where

Q_C

is heat removed from the cold reservoir and

W

is work input. First, calculate the heat removed:

Q_C = \text{COP} \times W = 3.2 \times 180 \text{ kJ} = 576 \text{ kJ}

. The total heat transferred to the environment (hot reservoir) includes both the heat removed from the cold space and the work energy input. By energy conservation:

Q_H = Q_C + W = 576 \text{ kJ} + 180 \text{ kJ} = 756 \text{ kJ}

. This confirms answer C is correct. Answer A (576 kJ) represents only the heat removed from the cold reservoir, missing the additional work energy that also becomes heat in the hot reservoir. Answer B (396 kJ) appears to subtract work from the cooling effect (576 - 180), which has no physical meaning in this context. Answer D (216 kJ) incorrectly adds the COP to the work (3.2 + 180), confusing the relationship between these quantities. Remember that in any heat engine or refrigerator cycle, energy is conserved. The hot reservoir always receives more energy than what's removed from the cold reservoir because the work input also ends up as heat in the hot reservoir. Always check that your answer satisfies

Q_H = Q_C + W

Question 12

A heat pump delivers $950 \text{ kJ}$ of heat to a house. If the COP of the heat pump is $4.75$ , how much heat is extracted from the cold reservoir?

750 kJ (correct answer)
950 kJ
200 kJ
1150 kJ
4512.5 kJ

Explanation: When you encounter heat pump problems, remember that a heat pump moves heat from a cold reservoir to a hot reservoir, and the coefficient of performance (COP) relates the desired output to the energy input required. For a heat pump,

\text{COP} = \frac{Q_H}{W}

, where

Q_H

is heat delivered to the hot reservoir (the house) and

W

is the work input. Since

\text{COP} = 4.75

and

Q_H = 950 \text{ kJ}

, you can find the work:

W = \frac{Q_H}{\text{COP}} = \frac{950}{4.75} = 200 \text{ kJ}

. Now apply the first law of thermodynamics. The heat delivered to the house equals the heat extracted from the cold reservoir plus the work input:

Q_H = Q_C + W

. Rearranging:

Q_C = Q_H - W = 950 - 200 = 750 \text{ kJ}

. Looking at the wrong answers: Choice B (950 kJ) incorrectly assumes the heat extracted equals the heat delivered, ignoring that work must be added. Choice C (200 kJ) gives you the work input, not the heat extracted - this is a common mix-up when students confuse what they've calculated. Choice D (1150 kJ) incorrectly adds work to the delivered heat, suggesting more energy comes out than goes in, which violates conservation of energy. The correct answer is A (750 kJ). Study tip: Always draw an energy flow diagram for heat pump problems. Mark what goes in (

Q_C + W

) and what comes out (

Q_H

), then apply conservation of energy. This visual approach prevents mixing up the different energy quantities.

Question 13

A refrigeration cycle operates with a compressor that consumes $5 \text{ kW}$ of power. If the refrigerator removes heat from the cold space at a rate of $18 \text{ kW}$ , what is the rate of heat rejection to the environment?

13 kW
18 kW
23 kW (correct answer)
5 kW
3.6 kW

Explanation: When you encounter refrigeration cycle problems, you're dealing with energy conservation—the first law of thermodynamics applied to heat engines and refrigerators. The key insight is that energy must be conserved: the total energy input equals the total energy output. In any refrigeration cycle, three energy flows are involved: work input to the compressor (

W

), heat removed from the cold space (

Q_C

), and heat rejected to the environment (

Q_H

). The first law requires that

Q_H = Q_C + W

. Think of it this way: the heat you remove from the cold space, plus the work energy you add via the compressor, must all be dumped to the hot environment. Here,

W = 5 \text{ kW}

and

Q_C = 18 \text{ kW}

, so

Q_H = 18 + 5 = 23 \text{ kW}

. This confirms answer C is correct. Looking at the wrong answers: A) 13 kW represents

Q_C - W

, which violates energy conservation—you can't reject less energy than you remove from the cold space. B) 18 kW ignores the compressor work entirely, assuming heat rejection equals heat removal, which would mean the compressor does nothing. D) 5 kW suggests only the work input is rejected, completely ignoring the removed heat. Remember this pattern: for any heat pump or refrigerator problem, always check that your energy flows satisfy

Q_H = Q_C + W

. The hot side always receives both the cold-side heat and the work input—energy cannot disappear.

Question 14

A heat pump system has a COP of $3.6$ and operates for $8 \text{ hours}$ while consuming $2.5 \text{ kW}$ of electrical power. How much heat energy is delivered to the heated space during this period?

20 kWh
72 kWh (correct answer)
52 kWh
28.8 kWh
92 kWh

Explanation: When you encounter heat pump problems, focus on understanding what the Coefficient of Performance (COP) tells you. The COP represents the ratio of useful heat delivered to the electrical energy input:

\text{COP} = \frac{Q_H}{W_{in}}

, where

Q_H

is heat delivered and

W_{in}

is work input. To find the heat delivered, rearrange this relationship:

Q_H = \text{COP} \times W_{in}

. First, calculate the total electrical energy consumed:

W_{in} = 2.5 \text{ kW} \times 8 \text{ hours} = 20 \text{ kWh}

. Then multiply by the COP:

Q_H = 3.6 \times 20 \text{ kWh} = 72 \text{ kWh}

. This confirms answer B is correct. Looking at the wrong answers: A (20 kWh) represents only the electrical energy input—this would be the answer if you forgot to multiply by the COP entirely. C (52 kWh) appears to come from subtracting the electrical input from the heat output (72 - 20 = 52), which might represent heat extracted from the outside air, but that's not what the question asks for. D (28.8 kWh) likely results from using the wrong time period or making an arithmetic error in the COP calculation. Remember that heat pumps are highly efficient because they move heat rather than generate it directly. The COP tells you how many units of heat you get per unit of electrical energy—always multiply your electrical input by the COP to find total heat delivered. Don't confuse this with heat extracted from the environment or net heat gain.

Question 15

A refrigerator removes $Q_L = 1400 \text{ kJ}$ from the cold reservoir during a cycle. If the COP of the refrigerator is $2.5$ , what is the heat rejected to the hot reservoir during the same cycle?

560 kJ
1400 kJ
1960 kJ (correct answer)
3500 kJ
840 kJ

Explanation: When you encounter refrigerator problems, remember that a refrigerator's job is to remove heat from a cold space and reject it to a hot space, requiring work input. The coefficient of performance (COP) relates these energy quantities. For a refrigerator,

\text{COP} = \frac{Q_L}{W}

, where

Q_L

is heat removed from the cold reservoir and

W

is work input. You can find the work:

W = \frac{Q_L}{\text{COP}} = \frac{1400}{2.5} = 560 \text{ kJ}

. Now apply energy conservation. The first law of thermodynamics requires that all energy input equals energy output:

Q_H = Q_L + W

, where

Q_H

is heat rejected to the hot reservoir. Therefore:

Q_H = 1400 + 560 = 1960 \text{ kJ}

. This confirms answer C is correct. Looking at the wrong answers: A) 560 kJ represents only the work input, not the total heat rejected. This misses the fact that the refrigerator must reject both the heat it removed plus the work energy. B) 1400 kJ is just the heat removed from the cold reservoir - this ignores the additional energy from work input that must also be rejected. D) 3500 kJ likely comes from incorrectly multiplying

Q_L \times \text{COP}

, which has no physical meaning in refrigerator cycles. Remember this pattern: for refrigerators, the heat rejected to the hot reservoir always equals the heat removed from the cold reservoir plus the work input. Energy must be conserved, so

Q_H = Q_L + W

is your key equation after finding work from the COP definition.

Question 16

Two identical heat pump systems operate in parallel, each with a COP of $4.8$ and each consuming $3 \text{ kW}$ of power. What is the total rate of heat delivery to the building?

14.4 kW
28.8 kW (correct answer)
6 kW
22.8 kW
57.6 kW

Explanation: When you encounter heat pump problems, remember that the coefficient of performance (COP) relates the useful heat output to the electrical power input. For heat pumps,

\text{COP} = \frac{\text{Heat delivered}}{\text{Power consumed}}

. Since you have two identical systems operating in parallel, you can analyze one system first, then double the result. For a single heat pump with COP = 4.8 and power consumption = 3 kW:

\text{Heat delivered per pump} = \text{COP} \times \text{Power consumed} = 4.8 \times 3 = 14.4 \text{ kW}

With two identical pumps operating simultaneously, the total heat delivery is:

\text{Total heat delivery} = 2 \times 14.4 = 28.8 \text{ kW}

Answer A (14.4 kW) represents the heat delivery from just one pump – you've forgotten to account for the second system. Answer C (6 kW) is simply the total electrical power consumed by both pumps (2 × 3 kW), not the heat delivered. Answer D (22.8 kW) might result from incorrectly calculating the COP relationship or making an arithmetic error in the multiplication. The key insight is that parallel operation means you add the individual outputs – each pump contributes its full capacity independently. Remember that COP values for heat pumps are typically greater than 1 because they move heat rather than generate it, so the heat output will always exceed the electrical input. Always double-check whether you're asked for individual or total system performance.

Question 17

A refrigerator cycle operates between reservoirs at $T_L = 250 \text{ K}$ and $T_H = 300 \text{ K}$ . If the actual COP is $60\%$ of the Carnot COP, what is the actual COP of this refrigerator?

3.0 (correct answer)
5.0
0.833
1.2
8.33

Explanation: When you encounter refrigerator cycle problems, you need to understand the relationship between ideal (Carnot) performance and real-world efficiency. Refrigerators are rated by their Coefficient of Performance (COP), which measures how much cooling they provide per unit of work input. First, calculate the Carnot COP, which represents the theoretical maximum performance. For a refrigerator operating between two thermal reservoirs, the Carnot COP is:

\text{COP}_{\text{Carnot}} = \frac{T_L}{T_H - T_L} = \frac{250}{300 - 250} = \frac{250}{50} = 5.0

Since the actual refrigerator operates at 60% of this ideal efficiency:

\text{COP}_{\text{actual}} = 0.60 \times 5.0 = 3.0

This confirms answer A is correct. Looking at the wrong answers: B (5.0) represents the Carnot COP itself - this trap catches students who forget to apply the 60% efficiency factor. C (0.833) likely comes from incorrectly calculating

\frac{T_H - T_L}{T_L}

and then applying the efficiency factor, confusing the refrigerator COP formula with the heat engine efficiency relationship. D (1.2) might result from calculation errors involving the temperature ratio or misapplying the efficiency percentage. Remember this pattern: always calculate the Carnot COP first using the temperature difference, then multiply by the given efficiency percentage. Real refrigerators never achieve Carnot performance, so your final answer should always be less than the theoretical maximum.

Question 18

A heat pump delivers $15 \text{ kW}$ of heating while extracting $12 \text{ kW}$ from the cold reservoir. If the electricity cost is $\$0.12 \text{ per kWh}$ , what is the hourly operating cost?

$1.80
$1.44
$0.36 (correct answer)
$3.24
$1.08

Explanation: When analyzing heat pump problems, you need to understand the energy flow: heat pumps use electrical work to move thermal energy from a cold reservoir to deliver heating to a warm space. The key insight is that you only pay for the electrical energy input, not the total heat delivered. To find the electrical work input, apply conservation of energy. The heat pump delivers 15 kW of heating while extracting 12 kW from the cold reservoir. Since energy must be conserved, the electrical work input equals the difference:

W = Q_H - Q_C = 15 \text{ kW} - 12 \text{ kW} = 3 \text{ kW}

The hourly operating cost is simply:

\text{Cost} = 3 \text{ kW} \times 1 \text{ hour} \times \$0.12/\text{kWh} = \$0.36

Let's examine why the other answers are incorrect. Answer A ($1.80) incorrectly uses the total heat delivered: $

15 \text{ kW} \times \

0.12 = $1.80

. This is wrong because you don't pay for heat extracted from the environment. Answer B ($1.44) mistakenly uses the heat extracted from the cold reservoir: $

12 \text{ kW} \times \

0.12 = $1.44

. Answer D ($3.24) appears to add both heat flows: $$(15 + 12) \times \$0.12 = \$3.24

, which double-counts energy and violates conservation principles. Remember this pattern: for heat pumps and refrigerators, always identify the electrical work input using energy conservation (

W = Q_H - Q_C

), then multiply by the electricity rate. Never use the total heat flows for cost calculations.

Question 19

A heat pump operating as both a heater in winter (COP = 3.8) and an air conditioner in summer (COP = 2.8) consumes the same power in both modes. What is the ratio of winter heat delivery to summer heat removal?

1.36 (correct answer)
0.737
1.0
10.64
6.6

Explanation: When you encounter heat pump problems, remember that COP (Coefficient of Performance) relates the useful energy output to the work input. For heating mode,

COP_h = \frac{Q_h}{W}

, and for cooling mode,

COP_c = \frac{Q_c}{W}

, where Q is heat transferred and W is work input. Since the heat pump consumes the same power (work input) in both modes, you can set up the ratio directly. The winter heat delivery is

Q_h = COP_h \times W = 3.8W

, and the summer heat removal is

Q_c = COP_c \times W = 2.8W

. Therefore, the ratio is:

\frac{Q_h}{Q_c} = \frac{3.8W}{2.8W} = \frac{3.8}{2.8} = 1.36

This confirms answer A is correct. Answer B (0.737) represents the inverse ratio - summer heat removal to winter heat delivery. This is a common mistake when students mix up which quantity goes in the numerator versus denominator. Answer C (1.0) would only be correct if both COP values were equal, meaning the heat pump performed identically in both modes. This ignores the given data showing different efficiencies. Answer D (10.64) appears to come from multiplying the COP values together (3.8 × 2.8) rather than taking their ratio. This represents a fundamental misunderstanding of how to compare the quantities. Study tip: In COP problems, always identify what stays constant (here, power consumption) and what varies (heat transfer). The ratio of outputs equals the ratio of COPs when input power is identical.

Question 20

A refrigerator operates between thermal reservoirs with a temperature difference of $30°C$ . If the cold reservoir temperature is $2°C$ , what is the maximum theoretical COP for this refrigerator?

9.17 (correct answer)
10.17
0.109
0.0917
8.17

Explanation: When you encounter refrigerator efficiency questions, you're dealing with the Coefficient of Performance (COP), which measures how effectively a refrigerator moves heat relative to the work input. The maximum theoretical COP occurs when the refrigerator operates as a Carnot refrigerator. For a Carnot refrigerator, the COP is given by:

COP = \frac{T_C}{T_H - T_C}

, where temperatures must be in Kelvin. First, convert the given temperatures:

T_C = 2°C + 273.15 = 275.15 K

and

T_H = 2°C + 30°C + 273.15 = 305.15 K

. Substituting into the formula:

COP = \frac{275.15}{305.15 - 275.15} = \frac{275.15}{30} = 9.17

Answer A (9.17) is correct because it properly applies the Carnot COP formula with temperatures converted to Kelvin. Answer B (10.17) likely results from incorrectly adding 1 to the calculated value, possibly confusing refrigerator COP with heat pump COP, which equals COP_refrigerator + 1. Answer C (0.109) appears to be the reciprocal of the correct answer, suggesting confusion between COP and efficiency formulas. Answer D (0.0917) is simply the correct answer divided by 100, possibly from a unit conversion error. Remember that COP for refrigerators is always greater than 1 for realistic temperature differences, and always convert temperatures to Kelvin before using Carnot formulas. Watch out for the common trap of confusing refrigerator and heat pump COP relationships.