QBManager

Thermodynamics

148639 A Carnot engine whose low-temperature reservoir is at $27^{\circ} C$ , has an efficiency $37.5 %$ . The high temperature reservoir is at

1

480^{\circ} C

2

327^{\circ} C

3

307^{\circ} C

4

207^{\circ} C

Explanation:

D Given, $T_{2} = 27 + 273 = 300 K$ and efficiency $(η) = 37.5 % = \frac{37.5}{100} = 0.375$
From Carnot equation of efficiency,
$η = 1 - \frac{T_{2}}{T_{1}}$
$0.375 = 1 - \frac{300}{T_{1}}$
$\frac{300}{T_{1}} = 1 - 0.375$
$\frac{300}{T_{1}} = 0.625$
$T_{1} = \frac{300}{0.625} = 480 K = 480 - 273 = 207^{\circ} C$

AP EAMCET-05.10.2021

Thermodynamics

148643 A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes $1 kWH$ of energy and the internal energy of the system drops by $5000 kJ$ . The quantity of heat transferred from the system to surrounding will be

1

- 8.6 MJ

2

86 MJ

3

- 86 MJ

4

- 8.6 KJ

Explanation:

A Given,
Power consumed regarded as work consumed which is taken as negative by sign convention
$∴ W = - 1 kWH = - 1 \times 3600 kJ = - 3600 kJ$
$Δ U = - 5000 kJ$
Applying first law of thermodynamics,
$Q - W = Δ U$
$Q - (- 3600) = - 5000$
$∴ Q = - 5000 - 3600$
$Q = - 8600 kJ$
$Q = - 8.6 MJ$

SRMJEEE-2019

Thermodynamics

148645 Inside the engine of an automobile, the cylinder compresses the air from approximately standard temperature and pressure to onetwentieth of the original volume and a pressure of $50 atm$ . What is the temperature of the compressed air?

1

500 K

2

682 K

3

550 K

4

1000 K

5

200 K

Explanation:

B Given, $P_{1} = 1 atm, P_{2} = 50 atm, V_{1} = V$ ,
$V_{2} = \frac{V}{20} & T_{1} = 273 K, T_{2} = ?$
From ideal gas equation-
$PV = n RT$
Taking ratio for two condition,
$\frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1 \times V}{50 \times \frac{V}{20}} = \frac{273}{T_{2}} \Rightarrow \frac{20}{50} = \frac{273}{T_{2}} \Rightarrow T_{2} = \frac{273 \times 5}{2}$
$T_{2} = 682 K$

Kerala CEE-2019

Thermodynamics

148646 An ideal mono atomic gas is taken round the cycle ABCDA as shown in figure. The work done during the cycle is :

1 PV

2

2 PV

3

\frac{P V}{2}

4 zero

Explanation:

A Work done in a process is equal to the area under the curve $P - V$ diagram

Work done during the cycle $=$ area of ABCDA
$W = BA \times DA$
$= (2 P - P) \times (2 V - V)$
$= PV$

IIT 1983

Thermodynamics

148639 A Carnot engine whose low-temperature reservoir is at $27^{\circ} C$ , has an efficiency $37.5 %$ . The high temperature reservoir is at

1

480^{\circ} C

2

327^{\circ} C

3

307^{\circ} C

4

207^{\circ} C

Explanation:

D Given, $T_{2} = 27 + 273 = 300 K$ and efficiency $(η) = 37.5 % = \frac{37.5}{100} = 0.375$
From Carnot equation of efficiency,
$η = 1 - \frac{T_{2}}{T_{1}}$
$0.375 = 1 - \frac{300}{T_{1}}$
$\frac{300}{T_{1}} = 1 - 0.375$
$\frac{300}{T_{1}} = 0.625$
$T_{1} = \frac{300}{0.625} = 480 K = 480 - 273 = 207^{\circ} C$

AP EAMCET-05.10.2021

Thermodynamics

148643 A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes $1 kWH$ of energy and the internal energy of the system drops by $5000 kJ$ . The quantity of heat transferred from the system to surrounding will be

1

- 8.6 MJ

2

86 MJ

3

- 86 MJ

4

- 8.6 KJ

Explanation:

A Given,
Power consumed regarded as work consumed which is taken as negative by sign convention
$∴ W = - 1 kWH = - 1 \times 3600 kJ = - 3600 kJ$
$Δ U = - 5000 kJ$
Applying first law of thermodynamics,
$Q - W = Δ U$
$Q - (- 3600) = - 5000$
$∴ Q = - 5000 - 3600$
$Q = - 8600 kJ$
$Q = - 8.6 MJ$

SRMJEEE-2019

Thermodynamics

148645 Inside the engine of an automobile, the cylinder compresses the air from approximately standard temperature and pressure to onetwentieth of the original volume and a pressure of $50 atm$ . What is the temperature of the compressed air?

1

500 K

2

682 K

3

550 K

4

1000 K

5

200 K

Explanation:

B Given, $P_{1} = 1 atm, P_{2} = 50 atm, V_{1} = V$ ,
$V_{2} = \frac{V}{20} & T_{1} = 273 K, T_{2} = ?$
From ideal gas equation-
$PV = n RT$
Taking ratio for two condition,
$\frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1 \times V}{50 \times \frac{V}{20}} = \frac{273}{T_{2}} \Rightarrow \frac{20}{50} = \frac{273}{T_{2}} \Rightarrow T_{2} = \frac{273 \times 5}{2}$
$T_{2} = 682 K$

Kerala CEE-2019

Thermodynamics

148646 An ideal mono atomic gas is taken round the cycle ABCDA as shown in figure. The work done during the cycle is :

1 PV

2

2 PV

3

\frac{P V}{2}

4 zero

Explanation:

A Work done in a process is equal to the area under the curve $P - V$ diagram

Work done during the cycle $=$ area of ABCDA
$W = BA \times DA$
$= (2 P - P) \times (2 V - V)$
$= PV$

IIT 1983

Thermodynamics

148639 A Carnot engine whose low-temperature reservoir is at $27^{\circ} C$ , has an efficiency $37.5 %$ . The high temperature reservoir is at

1

480^{\circ} C

2

327^{\circ} C

3

307^{\circ} C

4

207^{\circ} C

Explanation:

D Given, $T_{2} = 27 + 273 = 300 K$ and efficiency $(η) = 37.5 % = \frac{37.5}{100} = 0.375$
From Carnot equation of efficiency,
$η = 1 - \frac{T_{2}}{T_{1}}$
$0.375 = 1 - \frac{300}{T_{1}}$
$\frac{300}{T_{1}} = 1 - 0.375$
$\frac{300}{T_{1}} = 0.625$
$T_{1} = \frac{300}{0.625} = 480 K = 480 - 273 = 207^{\circ} C$

AP EAMCET-05.10.2021

Thermodynamics

148643 A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes $1 kWH$ of energy and the internal energy of the system drops by $5000 kJ$ . The quantity of heat transferred from the system to surrounding will be

1

- 8.6 MJ

2

86 MJ

3

- 86 MJ

4

- 8.6 KJ

Explanation:

A Given,
Power consumed regarded as work consumed which is taken as negative by sign convention
$∴ W = - 1 kWH = - 1 \times 3600 kJ = - 3600 kJ$
$Δ U = - 5000 kJ$
Applying first law of thermodynamics,
$Q - W = Δ U$
$Q - (- 3600) = - 5000$
$∴ Q = - 5000 - 3600$
$Q = - 8600 kJ$
$Q = - 8.6 MJ$

SRMJEEE-2019

Thermodynamics

148645 Inside the engine of an automobile, the cylinder compresses the air from approximately standard temperature and pressure to onetwentieth of the original volume and a pressure of $50 atm$ . What is the temperature of the compressed air?

1

500 K

2

682 K

3

550 K

4

1000 K

5

200 K

Explanation:

B Given, $P_{1} = 1 atm, P_{2} = 50 atm, V_{1} = V$ ,
$V_{2} = \frac{V}{20} & T_{1} = 273 K, T_{2} = ?$
From ideal gas equation-
$PV = n RT$
Taking ratio for two condition,
$\frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1 \times V}{50 \times \frac{V}{20}} = \frac{273}{T_{2}} \Rightarrow \frac{20}{50} = \frac{273}{T_{2}} \Rightarrow T_{2} = \frac{273 \times 5}{2}$
$T_{2} = 682 K$

Kerala CEE-2019

Thermodynamics

148646 An ideal mono atomic gas is taken round the cycle ABCDA as shown in figure. The work done during the cycle is :

1 PV

2

2 PV

3

\frac{P V}{2}

4 zero

Explanation:

A Work done in a process is equal to the area under the curve $P - V$ diagram

Work done during the cycle $=$ area of ABCDA
$W = BA \times DA$
$= (2 P - P) \times (2 V - V)$
$= PV$

IIT 1983

Thermodynamics

148639 A Carnot engine whose low-temperature reservoir is at $27^{\circ} C$ , has an efficiency $37.5 %$ . The high temperature reservoir is at

1

480^{\circ} C

2

327^{\circ} C

3

307^{\circ} C

4

207^{\circ} C

Explanation:

D Given, $T_{2} = 27 + 273 = 300 K$ and efficiency $(η) = 37.5 % = \frac{37.5}{100} = 0.375$
From Carnot equation of efficiency,
$η = 1 - \frac{T_{2}}{T_{1}}$
$0.375 = 1 - \frac{300}{T_{1}}$
$\frac{300}{T_{1}} = 1 - 0.375$
$\frac{300}{T_{1}} = 0.625$
$T_{1} = \frac{300}{0.625} = 480 K = 480 - 273 = 207^{\circ} C$

AP EAMCET-05.10.2021

Thermodynamics

148643 A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes $1 kWH$ of energy and the internal energy of the system drops by $5000 kJ$ . The quantity of heat transferred from the system to surrounding will be

1

- 8.6 MJ

2

86 MJ

3

- 86 MJ

4

- 8.6 KJ

Explanation:

A Given,
Power consumed regarded as work consumed which is taken as negative by sign convention
$∴ W = - 1 kWH = - 1 \times 3600 kJ = - 3600 kJ$
$Δ U = - 5000 kJ$
Applying first law of thermodynamics,
$Q - W = Δ U$
$Q - (- 3600) = - 5000$
$∴ Q = - 5000 - 3600$
$Q = - 8600 kJ$
$Q = - 8.6 MJ$

SRMJEEE-2019

Thermodynamics

148645 Inside the engine of an automobile, the cylinder compresses the air from approximately standard temperature and pressure to onetwentieth of the original volume and a pressure of $50 atm$ . What is the temperature of the compressed air?

1

500 K

2

682 K

3

550 K

4

1000 K

5

200 K

Explanation:

B Given, $P_{1} = 1 atm, P_{2} = 50 atm, V_{1} = V$ ,
$V_{2} = \frac{V}{20} & T_{1} = 273 K, T_{2} = ?$
From ideal gas equation-
$PV = n RT$
Taking ratio for two condition,
$\frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{1 \times V}{50 \times \frac{V}{20}} = \frac{273}{T_{2}} \Rightarrow \frac{20}{50} = \frac{273}{T_{2}} \Rightarrow T_{2} = \frac{273 \times 5}{2}$
$T_{2} = 682 K$

Kerala CEE-2019

Thermodynamics

148646 An ideal mono atomic gas is taken round the cycle ABCDA as shown in figure. The work done during the cycle is :

1 PV

2

2 PV

3

\frac{P V}{2}

4 zero

Explanation:

A Work done in a process is equal to the area under the curve $P - V$ diagram

Work done during the cycle $=$ area of ABCDA
$W = BA \times DA$
$= (2 P - P) \times (2 V - V)$
$= PV$

IIT 1983

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