A device operates between a hot reservoir at temperature $T_1$ (drawn at the top) and a cold reservoir at temperature $T_2$ (drawn at the bottom). In one cycle $Q_1$ is the heat added to the hot bath $T_1$ (the heat delivered by the device to $T_1$) and $Q_2$ is the heat taken from the cold bath $T_2$ (the heat drawn by the device from $T_2$). $W$ is the mechanical work done on the device. If $W>0$, which of the following are possible? More than one of the options may be correct. (a) $Q_1 > Q_2 > 0$ (b) $Q_2 > Q_1 > 0$ (c) $Q_2 0$

A device operates between a hot reservoir at temperature T₁ (drawn at the top) and a cold reservoir at tempera

NCERT Exemplar · Q11

T_1

(a)

Q_1 > Q_2 > 0

(b)

Q_2 > Q_1 > 0

(c)

Q_2 < Q_1 < 0

(d)

Q_1 < 0,\ Q_2 > 0

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Concept understanding — Refrigerator Coefficient of Performance

A refrigerator does not "create cold." It moves heat from a cold place (the inside of the fridge) to a hot place (the kitchen). That is the first thing to hold in your mind. The cold side gets colder because heat is being pulled out of it; the hot side gets hotter because that same heat is dumped there, plus the energy used to do the work.

If you put a hot bowl of soup inside the fridge, the fridge has to work harder. Why? Because more heat needs to be moved out. The "performance" of a fridge is simply: how much heat did I manage to pull out of the cold space, for every unit of work (electricity) I put in?

That is the intuition. You want a big "heat removed" for a small "work paid."

The precise statement

Let:

$Q_c$ = heat extracted from the cold reservoir (the inside of the fridge). This is the "useful" effect.
$Q_h$ = heat dumped into the hot reservoir (the kitchen). This is always larger than $Q_c$ .
$W$ = work input (electrical energy consumed by the compressor).

From the first law of thermodynamics (energy conservation), for a complete cycle:

Q_h = Q_c + W

The work you put in ends up as extra heat added to the hot side. So the Coefficient of Performance (COP) of a refrigerator is defined as:

\text{COP}_R = \frac{\text{What you want}}{\text{What you pay}} = \frac{Q_c}{W}

Using $W = Q_h - Q_c$ , we get the standard form:

\boxed{\text{COP}_R = \frac{Q_c}{Q_h - Q_c}}

Important

COP is not efficiency. Efficiency (for a heat engine) is always less than 1. COP for a refrigerator is always greater than 1 (often 2–6 for real fridges). Why? Because $Q_c$ can be several times larger than $W$ . You are moving heat, not converting it into work.

A concrete example

Suppose a fridge extracts 200 J of heat from the inside ( $Q_c = 200$ J) and dumps 250 J into the kitchen ( $Q_h = 250$ J). Then the work done is $W = 250 - 200 = 50$ J.

\text{COP} = \frac{200}{50} = 4

This means: for every 1 J of electrical energy you pay for, the fridge moves 4 J of heat out of your food. That is a COP of 4 — quite good.

If the fridge were perfect (impossible), it would move heat with zero work, and COP would be infinite. Real fridges have COP values between 2 and 6, depending on the temperature difference they have to work against.

Watch out

A common mistake: thinking COP = $Q_h/W$ or $Q_h/(Q_h - Q_c)$ . That is the COP of a heat pump (used for heating), not a refrigerator. For a fridge, the numerator is always $Q_c$ , the heat removed from the cold space.

Why the formula makes physical sense …

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With $Q_1$ the heat delivered to the hot bath, $Q_2$ the heat drawn from the cold bath and $W$ the work done on the device, conservation of energy over a cycle ( $\Delta U=0$ ) reads $Q_1=Q_2+W$ . Since $W>0$ , we need $Q_1>Q_2$ . That inequality holds both when both are positive (A) and when both are negative with $Q_1$ less negative (C).

Concept

Over one complete cycle the device returns to its initial state, so $\Delta U=0$ and the first law gives, for the stated sign convention (energy in $=$ energy out):

W+Q_2=Q_1\quad\Longrightarrow\quad Q_1-Q_2=W.

Applying $W>0$

$W>0\Rightarrow Q_1-Q_2>0\Rightarrow Q_1>Q_2$ . Now test the options:

(A) $Q_1>Q_2>0$ : satisfies $Q_1>Q_2$ — possible. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$-20^0C$
›Reveal solutionSolution
Uses the ideal (Carnot) coefficient of performance formula for a refrigerator to find the hot reservoir (surroundings) temperature from the given COP and freezer temperature. Answer: approximately $31^0C$ .

Concept and Intuition

A refrigerator pumps heat from a cold region (the freezer) to a hotter region (the surroundings), using work input. Its efficiency at this task is quantified by the coefficient of performance,
$COP = \frac{Q_c}{W} = \frac{T_c}{T_h-T_c}$
(the ideal/Carnot limit, using absolute temperatures). A higher COP means the refrigerator moves more heat per unit of work invested — and for a fixed cold-reservoir temperature, achieving a high COP requires the hot and cold reservoirs to be close in temperature (a small $T_h - T_c$ gap makes pumping heat "uphill" easier).

Step-by-Step Solution

Convert freezer temperature to Kelvin: $T_c = -20^0C = 273-20 = 253\ K$ .

Apply the COP formula: $COP = \dfrac{T_c}{T_h-T_c}$ , given $COP=5$ : $5 = \frac{253}{T_h - 253}$ …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$-13^\circ\mathrm{C}$
›Reveal solutionSolution
Using the coefficient-of-performance formula for a refrigerator with the freezer (cold reservoir) temperature given, the room (hot reservoir) temperature comes out to $39^\circ\mathrm{C}$ .

Concept and Intuition

A refrigerator moves heat from a cold reservoir (freezer, $T_2$ ) to a hot reservoir (room, $T_1$ ) using work input. For an ideal (Carnot) refrigerator, the coefficient of performance is $COP = \dfrac{Q_2}{W} = \dfrac{T_2}{T_1-T_2}$ , where both temperatures must be in Kelvin. A higher COP means less work is needed to extract a given amount of heat, and this relation lets us solve for the unknown reservoir temperature.

Step-by-Step Solution

Convert freezer temperature to Kelvin: $T_2 = -13+273 = 260\ \mathrm{K}$ .

Apply $COP = \dfrac{T_2}{T_1-T_2}$ with $COP=5$ : $5 = \dfrac{260}{T_1-260}$ .

$T_1 - 260 = \dfrac{260}{5} = 52$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$Q. If the coefficient of performance of a refrigerator is 5 and the temperature inside it is -20 °C, then the temperature of its surroundings is (A) 21.6 °C (B) 30.6 °C (C) 40.6 °C (D) 10.6 °C$
›Reveal solutionSolution
Using the ideal-refrigerator COP formula $\text{COP} = T_c/(T_h-T_c)$ with $T_c = 253$ K and COP = 5 gives a surrounding temperature of 30.6°C.

Concept and Intuition

A refrigerator moves heat from a cold reservoir (its interior) to a hot reservoir (the surroundings), consuming work in the process. Its coefficient of performance (COP) measures how much heat is extracted from the cold space per unit of work input, and for an ideal (Carnot) refrigerator this depends only on the absolute temperatures of the two reservoirs.

Step-by-Step Solution

Convert to Kelvin: $T_c = -20 + 273 = 253$ K.

Carnot refrigerator COP formula: $\text{COP} = \dfrac{T_c}{T_h - T_c}$ .

Given $\text{COP} = 5$ : $5 = \frac{253}{T_h - 253} \implies T_h - 253 = \frac{253}{5} = 50.6 \implies T_h = 303.6\ \text{K}$ …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$T_1$
›Reveal solutionSolution
This tests the standard Carnot expressions for engine efficiency and refrigerator COP between the same two reservoirs, and combining them algebraically.

Concept and Intuition

A heat engine extracts heat from the hot reservoir $T_1$ , does work, and rejects the rest to the cold reservoir $T_2$ ; its efficiency measures how much of the absorbed heat becomes useful work: $\eta = \dfrac{T_1-T_2}{T_1}$ .

A refrigerator does the reverse — work is put in to pump heat out of the cold reservoir $T_2$ into the hot one $T_1$ . Its coefficient of performance measures how much heat is removed from the cold side per unit work input: $\text{COP} = \dfrac{T_2}{T_1-T_2}$ .

Both quantities share the same reservoirs, so their ratio isolates how these two very different figures of merit relate for the same working temperatures.

Step-by-Step Solution

Efficiency of the engine: $\eta = \dfrac{T_1-T_2}{T_1}$ .

COP of the refrigerator: $\text{COP} = \dfrac{T_2}{T_1-T_2}$ . …
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AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$Q. A Carnot heat engine has an efficiency of 10 %. If the same engine is worked backward to obtain a refrigerator, then the coefficient of performance of the refrigerator is (A) 8 (B) 9 (C) 5 (D) 6$
›Reveal solutionSolution
The same reversible cycle run backward as a refrigerator has $\text{COP}=\dfrac{T_2}{T_1-T_2}$ ; using the engine's 10% efficiency to find the temperature ratio gives $\text{COP}=9$ .

Concept and Intuition

A Carnot engine and a Carnot refrigerator between the same two reservoirs are the same reversible cycle run in opposite directions, so their temperature relationships are linked. Knowing the engine's efficiency $\eta=1-T_2/T_1$ is enough to determine $T_2/T_1$ , and hence the refrigerator's coefficient of performance $\text{COP}=\dfrac{T_2}{T_1-T_2}$ , without needing the actual temperatures.

Step-by-Step Solution

Engine efficiency: $\eta=1-\dfrac{T_2}{T_1}=0.10\Rightarrow \dfrac{T_2}{T_1}=0.90$ .

So $T_1-T_2=T_1-0.9T_1=0.1T_1$ , and $T_2=0.9T_1$ .

Refrigerator COP $=\dfrac{T_2}{T_1-T_2}=\dfrac{0.9T_1}{0.1T_1}=9$ .

Common Mistakes …
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AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$^{-1}$
›Reveal solutionSolution
The refrigerator must continuously extract, at the cold end, exactly the heat that would otherwise melt the ice; combined with the Carnot COP between the ice-box and room temperatures, this gives the minimum motor power, about $13.6$ W.

Concept and Intuition

"Just prevents melting" means the refrigerator's rate of heat extraction from the cold reservoir (ice box) equals the rate at which heat leaks in from the warm room (which would otherwise melt ice at 2 kg/hr). The minimum power input for a given heat extraction rate is set by the ideal (Carnot) coefficient of performance between the two temperatures.

Step-by-Step Solution

Heat that must be removed per hour to stop 2 kg of ice from melting: $Q_1 = mL = 2000\text{ g}\times80\ \text{cal/g} = 160000$ cal/hr.

Convert to a rate in watts: $\dfrac{160000\times4.2\ \text{J}}{3600\ \text{s}} \approx 186.7$ W.

Cold reservoir temperature $T_C = 0°\text{C}=273$ K; hot reservoir (room) $T_H=20°\text{C}=293$ K.

Carnot COP (refrigerator): $\text{COP}=\dfrac{T_C}{T_H-T_C}=\dfrac{273}{20}=13.65$ . …
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AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
$\dfrac{100}{3}$
›Reveal solutionSolution
Using $COP=Q_c/W$ and energy conservation $Q_h=Q_c+W$ , the work done on the refrigerator's working substance is 200 J.

Concept and Intuition

A refrigerator extracts heat $Q_c$ from the cold reservoir using work $W$ done on it, and rejects $Q_h=Q_c+W$ to the hot reservoir (first law applied to a cyclic process). Its coefficient of performance is defined as $COP=\dfrac{Q_c}{W}$ (what you get — heat removed from the cold space — divided by what you pay — work input).

Step-by-Step Solution

Given: $COP=0.25$ , $Q_h=250$ J.

Energy conservation for the cycle: $Q_h=Q_c+W$ .

From the COP definition, $Q_c=COP\times W=0.25W$ .

Substitute into the energy balance: $Q_h=0.25W+W=1.25W$ .

So $W=\dfrac{Q_h}{1.25}=\dfrac{250}{1.25}=200$ J.

Common Mistakes …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$1/5$
›Reveal solutionSolution
A Carnot engine of efficiency $1/5$ run in reverse as a refrigerator has $COP=4$ ; with $50\ J$ of work input, it pumps $200\ J$ of heat out of the cold reservoir.

Concept and Intuition

A Carnot engine and a Carnot refrigerator operating between the same two temperatures are related — the same reversible cycle, run backwards, becomes a refrigerator. Its efficiency (as an engine) fixes the temperature ratio $T_2/T_1$ , which in turn fixes the coefficient of performance (COP) when the very same cycle is used as a refrigerator.

Step-by-Step Solution

Carnot efficiency as an engine: $\eta = 1-\dfrac{T_2}{T_1} = \dfrac{1}{5} \Rightarrow \dfrac{T_2}{T_1} = \dfrac{4}{5}$ .

Coefficient of performance as a refrigerator:

$COP = \frac{T_2}{T_1-T_2}$

Using $T_2/T_1 = 4/5$ , write $T_1 = 5k,\ T_2=4k$ for some constant $k$ : $COP = \frac{4k}{5k-4k} = \frac{4k}{k} = 4$ …
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Concept

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