JEE Advanced Carnot Engine, Refrigerators & Second Law of Thermodynamics PYQ

Q 1 :

A solid body of constant heat capacity $1 J / ° C$ is being heated by keeping it in contact with reservoirs in two ways : [2012]

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.

In both the cases body is brought from initial temperature $100 ° C$ to final temperature $200 ° C$ . Entropy change of the body in the two cases respectively is :

$l n 2, 2 l n 2$
$2 l n 2, 8 l n 2$
$l n 2, 4 l n 2$
$l n 2, l n 2$

1

2

3

4

(4)

The entropy change of the body in the two cases is same as entropy is a state function.

Q 2 :

The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4. It extracts 150 J of heat per cycle from the hot reservoir. The work extracted from this engine is being fully used to run a heat pump which has a coefficient of performance 10. The hot reservoir of the heat pump is at a temperature of 300 K. Which of the following statements is/are correct: [2025]

Work extracted from the Carnot engine in one cycle is 60 J.
Temperature of the cold reservoir of the Carnot engine is 600 K.
Temperature of the cold reservoir of the heat pump is 270 K.
Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.

1

2

3

4

Select one or more options

(1, 2, 3)

[IMAGE 498]

$Efficiency of engine η = \frac{W}{Q_{1}} and coefficient of performance (COP) of heat pump = \frac{Q_{3}}{W}$

$η = 0.4$

${(COP)}_{HP} = 10$

$W = η \times Q_{1} = 0.4 \times 150 = 60 J$

$So, option (1) is correct.$

$10 = \frac{Q_{3}}{W} = \frac{Q_{3}}{60}$

$∴$ $Q_{3} = 600 J$

$So option (2) is correct.$

$Also η = 0.4 = 1 - \frac{T_{2}}{T_{1}}$

$T_{2} = 600 K$

$COP = \frac{T_{3}}{T_{3} - T_{4}} = 10 = \frac{300}{300 - T_{4}} = 10$

$∴$ $T_{4} = 270 K$

$So option (3) is correct.$

Q 3 :

A fixed thermally conducting cylinder has a radius R and height $L_{0}$ . The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is $P_{0}$ .

[IMAGE 499]

Q. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be [2007]

$P_{0}$
$\frac{P_{0}}{2}$
$\frac{P_{0}}{2} + \frac{M g}{π R^{2}}$
$\frac{P_{0}}{2} - \frac{M g}{π R^{2}}$

1

2

3

4

(1)

As the cylinder is open at its bottom and has a small hole at its top, hence the pressure inside the cylinder is atmospheric pressure, $P_{0}$ throughout the slow pulling process.

Q 4 :

A fixed thermally conducting cylinder has a radius R and height $L_{0}$ . The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is $P_{0}$ .

[IMAGE 500]

Q. While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is [2007]

$(\frac{2 P_{0} π R^{2}}{π R^{2} P_{0} + M g}) (2 L)$
$(\frac{P_{0} π R^{2} - M g}{π R^{2} P_{0}}) (2 L)$
$(\frac{P_{0} π R^{2} + M g}{π R^{2} P_{0}}) (2 L)$
$(\frac{P_{0} π R^{2}}{π R^{2} P_{0} - M g}) (2 L)$

1

2

3

4

(4)

[IMAGE 501]

$Let x be the distance of the piston from the top.$

$At equilibrium$

$M g = (P_{0} - p) π R^{2}$

$\Rightarrow p = \frac{- M g}{π R^{2}} + P_{0}$

$Since the cylinder is isothermally conducting$

$∴$ $temperature, T = constant$

$Applying P_{1} V_{1} = P_{2} V_{2}$

$P_{0} \times (2 L \times π R^{2}) = p \times x \times π R^{2}$ $\Rightarrow x = \frac{P_{0}}{p} \times 2 L$

$= \frac{P_{0}}{[P_{0} - \frac{M g}{π R^{2}}]} \times 2 L$

$= [\frac{P_{0} π R^{2}}{P_{0} π R^{2} - M g}] \times 2 L$

Q 5 :

A fixed thermally conducting cylinder has a radius R and height $L_{0}$ . The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is $P_{0}$ .

[IMAGE 502]

Q. The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $ρ$ . In equilibrium, the height H of the water column in the cylinder satisfies [2007]

$ρ g {(L_{0} - H)}^{2} + P_{0} (L_{0} - H) + L_{0} P_{0} = 0$
$ρ g {(L_{0} - H)}^{2} - P_{0} (L_{0} - H) - L_{0} P_{0} = 0$
$ρ g {(L_{0} - H)}^{2} + P_{0} (L_{0} - H) - L_{0} P_{0} = 0$
$ρ g {(L_{0} - H)}^{2} - P_{0} (L_{0} - H) + L_{0} P_{0} = 0$

1

2

3

4

(3)

[IMAGE 503]

$At equilibrium, p = P$

$\Rightarrow p = P_{0} + (L_{0} - H) ρ g ...(i)$

$Also P_{0} \times (π R^{2} L_{0}) = p [π R^{2} (L_{0} - H)]$ $(∵ P_{1} V_{1} = P_{2} V_{2})$

$\Rightarrow p = \frac{L_{0} P_{0}}{L_{0} - H} ...(ii)$

$Substituting this value of p from (i) and (ii)$

$\frac{L_{0} P_{0}}{L_{0} - H} = P_{0} + (L_{0} - H) ρ g$

$\Rightarrow L_{0} P_{0} = P_{0} (L_{0} - H) + {(L_{0} - H)}^{2} ρ g$

$\Rightarrow ρ g {(L_{0} - H)}^{2} + P_{0} (L_{0} - H) - L_{0} P_{0} = 0$

Topic Question Set

Want Us to Email you About Special Offers & Updates?