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Thermodynamics

148205 $50 g$ ice at $0^{\circ} C$ in insulator vessel, $50 g$ water of $100^{\circ} C$ is mixed in it, then final temperature of the mixture is (neglect the heat loss)

1

10^{\circ} C

2

0^{\circ} <<< T_{m} < 20^{\circ} C

3

20^{\circ} C

4 above

20^{\circ} C

Explanation:

A Let the final temperature be $T$ .
Heat released at $100^{\circ} C$ of water $Q_{1} = m_{1} s Δ t$
Heat gained by $0^{\circ} C$ ice to convert at $T^{\circ} C$ of water
$Q_{2} = m_{2} L + m_{2} s Δ t$
According to calorimetric principle,
Heat lost by $100^{\circ} C$ of water $=$ Heat gain by $0^{\circ} C$ of ice
$Q_{1} = Q_{2}$
$m_{1} s Δ t = m_{2} s Δ t + m_{2} L$
$50 \times 1 \times (100 - T) = 50 \times 1 \times (T - 0) + 50 \times 80$
${\begin{cases} s = 1 \frac{cal}{gm} for water \\ L = 80 \frac{cal}{gm} for ice \end{cases}}$
$100 - T = T + 80$
$100 - 80 = 2 T$
$20 = 2 T$
$T = 10^{\circ} C$

JIPMER-2014

Thermodynamics

148206 Heat energy absorbed by a system in going through a cyclic process shown in figure. The work done during the process is:

1

10^{7} π J

2

10^{4} π J

3

10^{3} π J

4

10^{2} π J

Explanation:

D $Δ U = 0$
$Δ W =$ Area of loop
$= π \times 10 \times 10 \times 10^{- 3} \times 10^{3} = 100 π$
$Δ Q = Δ U + Δ W = 10^{2} π$

Assam CEE-2017

Thermodynamics

148207 The Entropy (S) of a black hole can be written as $S = β k_{B} A$ , where $k_{B}$ is the Boltzmann constant and $A$ is the area of the black hole. Then $β$ has dimension of

1

L^{2}

2

M L^{2} T^{- 1}

3

L^{- 2}

4 dimensionless

Explanation:

C Entorpy, $S = β K_{B} A \dots \dots \dots$ . (i)
Dimensional unit of entropy
$S = [{ML}^{2} T^{- 2} K^{- 1}]$
Boltzmann constant, $K_{B} = [{ML}^{2} T^{- 2} K^{- 1}]$
Area $(A) = (L^{2})$
From equation. (i)
$[{ML}^{2} T^{- 2} K^{- 1}] = β [{ML}^{2} T^{- 2} K^{- 1}] [L^{2}]$
$β = \frac{[{ML}^{2} T^{- 2} K^{- 1}]}{[{ML}^{4} T^{- 2} K^{- 1}]} = [\frac{1}{L^{2}}]$
$β = [L^{- 2}]$
Hence option (c) is correct.

WB JEE 2022

Thermodynamics

148209 Consider a ball of mass $100 g$ attached to one end of a spring $(k = 800 N / m)$ and immersed in $0.5 kg$ water. Assume the complete system is in thermal equilibrium. The spring is now stretched to $20 cm$ and the mass is released so that it vibrates up and down. Estimate the change in temperature of water before the vibrations stop.
(Specific heat of the material of the ball $= 400$ $J / kg K$ specific heat of water $= 4200 J / kg / K$ )

1

8.2 \times 10^{- 4} K

2

10^{- 3} K

3

7.5 \times 10^{- 3} K

4

10^{- 3} K

Explanation:

C Given that,
Mass of ball $(m_{b}) = 100 gm$
spring constant $(k) = 800 N / m$
Stretched in spring $(x) = 20 cm$
Specific heat of the material of ball (c) $= 400 J / kg / K$
Specific heat of water $(c_{v}) = 4200 J / kg / K$
Energy, $E = \frac{1}{2} {kx}^{2}$
$E = \frac{1}{2} \times 800 \times {(20 \times 10^{- 2})}^{2}$
$= 16 Joule$
In water, $Q = m_{v} Δ T + m_{b} c Δ T$
$= 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 2140 Δ T$
$Δ T = \frac{16}{2140}$
$= 0.00747$
$= 7.5 \times 10^{- 3} K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148210 A monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be:

1

P

2

8 P

3

32 P

4

64 P

Explanation:

C Given that,
Initial Pressure $= P_{1}$
Initial volume $= V_{1}$
Compressed Volume $(V_{2}) = \frac{V_{1}}{8}$
As we know that
${PV}^{γ} = c$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ} \Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2} / 8})}^{γ}$
$γ = \frac{5}{3}$ for monoatomic gas.
$\frac{P_{2}}{P_{1}} = {(\frac{8}{1})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{5 / 3}$
$P_{2} / P_{1} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$
$P_{2} = 32 P_{1}$

JEE Main-26.07.2022

Thermodynamics

148205 $50 g$ ice at $0^{\circ} C$ in insulator vessel, $50 g$ water of $100^{\circ} C$ is mixed in it, then final temperature of the mixture is (neglect the heat loss)

1

10^{\circ} C

2

0^{\circ} <<< T_{m} < 20^{\circ} C

3

20^{\circ} C

4 above

20^{\circ} C

Explanation:

A Let the final temperature be $T$ .
Heat released at $100^{\circ} C$ of water $Q_{1} = m_{1} s Δ t$
Heat gained by $0^{\circ} C$ ice to convert at $T^{\circ} C$ of water
$Q_{2} = m_{2} L + m_{2} s Δ t$
According to calorimetric principle,
Heat lost by $100^{\circ} C$ of water $=$ Heat gain by $0^{\circ} C$ of ice
$Q_{1} = Q_{2}$
$m_{1} s Δ t = m_{2} s Δ t + m_{2} L$
$50 \times 1 \times (100 - T) = 50 \times 1 \times (T - 0) + 50 \times 80$
${\begin{cases} s = 1 \frac{cal}{gm} for water \\ L = 80 \frac{cal}{gm} for ice \end{cases}}$
$100 - T = T + 80$
$100 - 80 = 2 T$
$20 = 2 T$
$T = 10^{\circ} C$

JIPMER-2014

Thermodynamics

148206 Heat energy absorbed by a system in going through a cyclic process shown in figure. The work done during the process is:

1

10^{7} π J

2

10^{4} π J

3

10^{3} π J

4

10^{2} π J

Explanation:

D $Δ U = 0$
$Δ W =$ Area of loop
$= π \times 10 \times 10 \times 10^{- 3} \times 10^{3} = 100 π$
$Δ Q = Δ U + Δ W = 10^{2} π$

Assam CEE-2017

Thermodynamics

148207 The Entropy (S) of a black hole can be written as $S = β k_{B} A$ , where $k_{B}$ is the Boltzmann constant and $A$ is the area of the black hole. Then $β$ has dimension of

1

L^{2}

2

M L^{2} T^{- 1}

3

L^{- 2}

4 dimensionless

Explanation:

C Entorpy, $S = β K_{B} A \dots \dots \dots$ . (i)
Dimensional unit of entropy
$S = [{ML}^{2} T^{- 2} K^{- 1}]$
Boltzmann constant, $K_{B} = [{ML}^{2} T^{- 2} K^{- 1}]$
Area $(A) = (L^{2})$
From equation. (i)
$[{ML}^{2} T^{- 2} K^{- 1}] = β [{ML}^{2} T^{- 2} K^{- 1}] [L^{2}]$
$β = \frac{[{ML}^{2} T^{- 2} K^{- 1}]}{[{ML}^{4} T^{- 2} K^{- 1}]} = [\frac{1}{L^{2}}]$
$β = [L^{- 2}]$
Hence option (c) is correct.

WB JEE 2022

Thermodynamics

148209 Consider a ball of mass $100 g$ attached to one end of a spring $(k = 800 N / m)$ and immersed in $0.5 kg$ water. Assume the complete system is in thermal equilibrium. The spring is now stretched to $20 cm$ and the mass is released so that it vibrates up and down. Estimate the change in temperature of water before the vibrations stop.
(Specific heat of the material of the ball $= 400$ $J / kg K$ specific heat of water $= 4200 J / kg / K$ )

1

8.2 \times 10^{- 4} K

2

10^{- 3} K

3

7.5 \times 10^{- 3} K

4

10^{- 3} K

Explanation:

C Given that,
Mass of ball $(m_{b}) = 100 gm$
spring constant $(k) = 800 N / m$
Stretched in spring $(x) = 20 cm$
Specific heat of the material of ball (c) $= 400 J / kg / K$
Specific heat of water $(c_{v}) = 4200 J / kg / K$
Energy, $E = \frac{1}{2} {kx}^{2}$
$E = \frac{1}{2} \times 800 \times {(20 \times 10^{- 2})}^{2}$
$= 16 Joule$
In water, $Q = m_{v} Δ T + m_{b} c Δ T$
$= 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 2140 Δ T$
$Δ T = \frac{16}{2140}$
$= 0.00747$
$= 7.5 \times 10^{- 3} K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148210 A monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be:

1

P

2

8 P

3

32 P

4

64 P

Explanation:

C Given that,
Initial Pressure $= P_{1}$
Initial volume $= V_{1}$
Compressed Volume $(V_{2}) = \frac{V_{1}}{8}$
As we know that
${PV}^{γ} = c$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ} \Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2} / 8})}^{γ}$
$γ = \frac{5}{3}$ for monoatomic gas.
$\frac{P_{2}}{P_{1}} = {(\frac{8}{1})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{5 / 3}$
$P_{2} / P_{1} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$
$P_{2} = 32 P_{1}$

JEE Main-26.07.2022

Thermodynamics

148205 $50 g$ ice at $0^{\circ} C$ in insulator vessel, $50 g$ water of $100^{\circ} C$ is mixed in it, then final temperature of the mixture is (neglect the heat loss)

1

10^{\circ} C

2

0^{\circ} <<< T_{m} < 20^{\circ} C

3

20^{\circ} C

4 above

20^{\circ} C

Explanation:

A Let the final temperature be $T$ .
Heat released at $100^{\circ} C$ of water $Q_{1} = m_{1} s Δ t$
Heat gained by $0^{\circ} C$ ice to convert at $T^{\circ} C$ of water
$Q_{2} = m_{2} L + m_{2} s Δ t$
According to calorimetric principle,
Heat lost by $100^{\circ} C$ of water $=$ Heat gain by $0^{\circ} C$ of ice
$Q_{1} = Q_{2}$
$m_{1} s Δ t = m_{2} s Δ t + m_{2} L$
$50 \times 1 \times (100 - T) = 50 \times 1 \times (T - 0) + 50 \times 80$
${\begin{cases} s = 1 \frac{cal}{gm} for water \\ L = 80 \frac{cal}{gm} for ice \end{cases}}$
$100 - T = T + 80$
$100 - 80 = 2 T$
$20 = 2 T$
$T = 10^{\circ} C$

JIPMER-2014

Thermodynamics

148206 Heat energy absorbed by a system in going through a cyclic process shown in figure. The work done during the process is:

1

10^{7} π J

2

10^{4} π J

3

10^{3} π J

4

10^{2} π J

Explanation:

D $Δ U = 0$
$Δ W =$ Area of loop
$= π \times 10 \times 10 \times 10^{- 3} \times 10^{3} = 100 π$
$Δ Q = Δ U + Δ W = 10^{2} π$

Assam CEE-2017

Thermodynamics

148207 The Entropy (S) of a black hole can be written as $S = β k_{B} A$ , where $k_{B}$ is the Boltzmann constant and $A$ is the area of the black hole. Then $β$ has dimension of

1

L^{2}

2

M L^{2} T^{- 1}

3

L^{- 2}

4 dimensionless

Explanation:

C Entorpy, $S = β K_{B} A \dots \dots \dots$ . (i)
Dimensional unit of entropy
$S = [{ML}^{2} T^{- 2} K^{- 1}]$
Boltzmann constant, $K_{B} = [{ML}^{2} T^{- 2} K^{- 1}]$
Area $(A) = (L^{2})$
From equation. (i)
$[{ML}^{2} T^{- 2} K^{- 1}] = β [{ML}^{2} T^{- 2} K^{- 1}] [L^{2}]$
$β = \frac{[{ML}^{2} T^{- 2} K^{- 1}]}{[{ML}^{4} T^{- 2} K^{- 1}]} = [\frac{1}{L^{2}}]$
$β = [L^{- 2}]$
Hence option (c) is correct.

WB JEE 2022

Thermodynamics

148209 Consider a ball of mass $100 g$ attached to one end of a spring $(k = 800 N / m)$ and immersed in $0.5 kg$ water. Assume the complete system is in thermal equilibrium. The spring is now stretched to $20 cm$ and the mass is released so that it vibrates up and down. Estimate the change in temperature of water before the vibrations stop.
(Specific heat of the material of the ball $= 400$ $J / kg K$ specific heat of water $= 4200 J / kg / K$ )

1

8.2 \times 10^{- 4} K

2

10^{- 3} K

3

7.5 \times 10^{- 3} K

4

10^{- 3} K

Explanation:

C Given that,
Mass of ball $(m_{b}) = 100 gm$
spring constant $(k) = 800 N / m$
Stretched in spring $(x) = 20 cm$
Specific heat of the material of ball (c) $= 400 J / kg / K$
Specific heat of water $(c_{v}) = 4200 J / kg / K$
Energy, $E = \frac{1}{2} {kx}^{2}$
$E = \frac{1}{2} \times 800 \times {(20 \times 10^{- 2})}^{2}$
$= 16 Joule$
In water, $Q = m_{v} Δ T + m_{b} c Δ T$
$= 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 2140 Δ T$
$Δ T = \frac{16}{2140}$
$= 0.00747$
$= 7.5 \times 10^{- 3} K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148210 A monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be:

1

P

2

8 P

3

32 P

4

64 P

Explanation:

C Given that,
Initial Pressure $= P_{1}$
Initial volume $= V_{1}$
Compressed Volume $(V_{2}) = \frac{V_{1}}{8}$
As we know that
${PV}^{γ} = c$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ} \Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2} / 8})}^{γ}$
$γ = \frac{5}{3}$ for monoatomic gas.
$\frac{P_{2}}{P_{1}} = {(\frac{8}{1})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{5 / 3}$
$P_{2} / P_{1} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$
$P_{2} = 32 P_{1}$

JEE Main-26.07.2022

Thermodynamics

148205 $50 g$ ice at $0^{\circ} C$ in insulator vessel, $50 g$ water of $100^{\circ} C$ is mixed in it, then final temperature of the mixture is (neglect the heat loss)

1

10^{\circ} C

2

0^{\circ} <<< T_{m} < 20^{\circ} C

3

20^{\circ} C

4 above

20^{\circ} C

Explanation:

A Let the final temperature be $T$ .
Heat released at $100^{\circ} C$ of water $Q_{1} = m_{1} s Δ t$
Heat gained by $0^{\circ} C$ ice to convert at $T^{\circ} C$ of water
$Q_{2} = m_{2} L + m_{2} s Δ t$
According to calorimetric principle,
Heat lost by $100^{\circ} C$ of water $=$ Heat gain by $0^{\circ} C$ of ice
$Q_{1} = Q_{2}$
$m_{1} s Δ t = m_{2} s Δ t + m_{2} L$
$50 \times 1 \times (100 - T) = 50 \times 1 \times (T - 0) + 50 \times 80$
${\begin{cases} s = 1 \frac{cal}{gm} for water \\ L = 80 \frac{cal}{gm} for ice \end{cases}}$
$100 - T = T + 80$
$100 - 80 = 2 T$
$20 = 2 T$
$T = 10^{\circ} C$

JIPMER-2014

Thermodynamics

148206 Heat energy absorbed by a system in going through a cyclic process shown in figure. The work done during the process is:

1

10^{7} π J

2

10^{4} π J

3

10^{3} π J

4

10^{2} π J

Explanation:

D $Δ U = 0$
$Δ W =$ Area of loop
$= π \times 10 \times 10 \times 10^{- 3} \times 10^{3} = 100 π$
$Δ Q = Δ U + Δ W = 10^{2} π$

Assam CEE-2017

Thermodynamics

148207 The Entropy (S) of a black hole can be written as $S = β k_{B} A$ , where $k_{B}$ is the Boltzmann constant and $A$ is the area of the black hole. Then $β$ has dimension of

1

L^{2}

2

M L^{2} T^{- 1}

3

L^{- 2}

4 dimensionless

Explanation:

C Entorpy, $S = β K_{B} A \dots \dots \dots$ . (i)
Dimensional unit of entropy
$S = [{ML}^{2} T^{- 2} K^{- 1}]$
Boltzmann constant, $K_{B} = [{ML}^{2} T^{- 2} K^{- 1}]$
Area $(A) = (L^{2})$
From equation. (i)
$[{ML}^{2} T^{- 2} K^{- 1}] = β [{ML}^{2} T^{- 2} K^{- 1}] [L^{2}]$
$β = \frac{[{ML}^{2} T^{- 2} K^{- 1}]}{[{ML}^{4} T^{- 2} K^{- 1}]} = [\frac{1}{L^{2}}]$
$β = [L^{- 2}]$
Hence option (c) is correct.

WB JEE 2022

Thermodynamics

148209 Consider a ball of mass $100 g$ attached to one end of a spring $(k = 800 N / m)$ and immersed in $0.5 kg$ water. Assume the complete system is in thermal equilibrium. The spring is now stretched to $20 cm$ and the mass is released so that it vibrates up and down. Estimate the change in temperature of water before the vibrations stop.
(Specific heat of the material of the ball $= 400$ $J / kg K$ specific heat of water $= 4200 J / kg / K$ )

1

8.2 \times 10^{- 4} K

2

10^{- 3} K

3

7.5 \times 10^{- 3} K

4

10^{- 3} K

Explanation:

C Given that,
Mass of ball $(m_{b}) = 100 gm$
spring constant $(k) = 800 N / m$
Stretched in spring $(x) = 20 cm$
Specific heat of the material of ball (c) $= 400 J / kg / K$
Specific heat of water $(c_{v}) = 4200 J / kg / K$
Energy, $E = \frac{1}{2} {kx}^{2}$
$E = \frac{1}{2} \times 800 \times {(20 \times 10^{- 2})}^{2}$
$= 16 Joule$
In water, $Q = m_{v} Δ T + m_{b} c Δ T$
$= 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 2140 Δ T$
$Δ T = \frac{16}{2140}$
$= 0.00747$
$= 7.5 \times 10^{- 3} K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148210 A monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be:

1

P

2

8 P

3

32 P

4

64 P

Explanation:

C Given that,
Initial Pressure $= P_{1}$
Initial volume $= V_{1}$
Compressed Volume $(V_{2}) = \frac{V_{1}}{8}$
As we know that
${PV}^{γ} = c$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ} \Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2} / 8})}^{γ}$
$γ = \frac{5}{3}$ for monoatomic gas.
$\frac{P_{2}}{P_{1}} = {(\frac{8}{1})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{5 / 3}$
$P_{2} / P_{1} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$
$P_{2} = 32 P_{1}$

JEE Main-26.07.2022

Thermodynamics

148205 $50 g$ ice at $0^{\circ} C$ in insulator vessel, $50 g$ water of $100^{\circ} C$ is mixed in it, then final temperature of the mixture is (neglect the heat loss)

1

10^{\circ} C

2

0^{\circ} <<< T_{m} < 20^{\circ} C

3

20^{\circ} C

4 above

20^{\circ} C

Explanation:

A Let the final temperature be $T$ .
Heat released at $100^{\circ} C$ of water $Q_{1} = m_{1} s Δ t$
Heat gained by $0^{\circ} C$ ice to convert at $T^{\circ} C$ of water
$Q_{2} = m_{2} L + m_{2} s Δ t$
According to calorimetric principle,
Heat lost by $100^{\circ} C$ of water $=$ Heat gain by $0^{\circ} C$ of ice
$Q_{1} = Q_{2}$
$m_{1} s Δ t = m_{2} s Δ t + m_{2} L$
$50 \times 1 \times (100 - T) = 50 \times 1 \times (T - 0) + 50 \times 80$
${\begin{cases} s = 1 \frac{cal}{gm} for water \\ L = 80 \frac{cal}{gm} for ice \end{cases}}$
$100 - T = T + 80$
$100 - 80 = 2 T$
$20 = 2 T$
$T = 10^{\circ} C$

JIPMER-2014

Thermodynamics

148206 Heat energy absorbed by a system in going through a cyclic process shown in figure. The work done during the process is:

1

10^{7} π J

2

10^{4} π J

3

10^{3} π J

4

10^{2} π J

Explanation:

D $Δ U = 0$
$Δ W =$ Area of loop
$= π \times 10 \times 10 \times 10^{- 3} \times 10^{3} = 100 π$
$Δ Q = Δ U + Δ W = 10^{2} π$

Assam CEE-2017

Thermodynamics

148207 The Entropy (S) of a black hole can be written as $S = β k_{B} A$ , where $k_{B}$ is the Boltzmann constant and $A$ is the area of the black hole. Then $β$ has dimension of

1

L^{2}

2

M L^{2} T^{- 1}

3

L^{- 2}

4 dimensionless

Explanation:

C Entorpy, $S = β K_{B} A \dots \dots \dots$ . (i)
Dimensional unit of entropy
$S = [{ML}^{2} T^{- 2} K^{- 1}]$
Boltzmann constant, $K_{B} = [{ML}^{2} T^{- 2} K^{- 1}]$
Area $(A) = (L^{2})$
From equation. (i)
$[{ML}^{2} T^{- 2} K^{- 1}] = β [{ML}^{2} T^{- 2} K^{- 1}] [L^{2}]$
$β = \frac{[{ML}^{2} T^{- 2} K^{- 1}]}{[{ML}^{4} T^{- 2} K^{- 1}]} = [\frac{1}{L^{2}}]$
$β = [L^{- 2}]$
Hence option (c) is correct.

WB JEE 2022

Thermodynamics

148209 Consider a ball of mass $100 g$ attached to one end of a spring $(k = 800 N / m)$ and immersed in $0.5 kg$ water. Assume the complete system is in thermal equilibrium. The spring is now stretched to $20 cm$ and the mass is released so that it vibrates up and down. Estimate the change in temperature of water before the vibrations stop.
(Specific heat of the material of the ball $= 400$ $J / kg K$ specific heat of water $= 4200 J / kg / K$ )

1

8.2 \times 10^{- 4} K

2

10^{- 3} K

3

7.5 \times 10^{- 3} K

4

10^{- 3} K

Explanation:

C Given that,
Mass of ball $(m_{b}) = 100 gm$
spring constant $(k) = 800 N / m$
Stretched in spring $(x) = 20 cm$
Specific heat of the material of ball (c) $= 400 J / kg / K$
Specific heat of water $(c_{v}) = 4200 J / kg / K$
Energy, $E = \frac{1}{2} {kx}^{2}$
$E = \frac{1}{2} \times 800 \times {(20 \times 10^{- 2})}^{2}$
$= 16 Joule$
In water, $Q = m_{v} Δ T + m_{b} c Δ T$
$= 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 0.5 \times 4200 \times Δ T + 0.1 \times 400 \times Δ T$
$16 = 2140 Δ T$
$Δ T = \frac{16}{2140}$
$= 0.00747$
$= 7.5 \times 10^{- 3} K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148210 A monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be:

1

P

2

8 P

3

32 P

4

64 P

Explanation:

C Given that,
Initial Pressure $= P_{1}$
Initial volume $= V_{1}$
Compressed Volume $(V_{2}) = \frac{V_{1}}{8}$
As we know that
${PV}^{γ} = c$
$\frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ} \Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{V_{1}}{V_{2} / 8})}^{γ}$
$γ = \frac{5}{3}$ for monoatomic gas.
$\frac{P_{2}}{P_{1}} = {(\frac{8}{1})}^{5 / 3}$
$\frac{P_{2}}{P_{1}} = {(2^{3})}^{5 / 3}$
$P_{2} / P_{1} = 2^{5}$
$\frac{P_{2}}{P_{1}} = 32$
$P_{2} = 32 P_{1}$

JEE Main-26.07.2022

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Question Bank Series

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