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

Punjab PsebMCQ· 1mImportance★★★★★est

54% · 19/35 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ10 — An engine is proposed to take one mole of an ideal gas once around the…→Next questionQ12 — Can a system be heated and its temperature remains constant?

Concept

Applying W>0W>0W>0

🎓Unlock everything free for 14 days

Applying $W>0$

Unlock everything free for 14 days