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Thermodynamics

148540 An electrical refrigerator with $β = 5$ extracts $5000 J$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.

1

1 kJ

2

0.5 kJ

3

0.8 kJ

4

1.2 kJ

Explanation:

A Given,
Coefficient of performance of refrigerator-
$β = 5$
Amount of heat removed $(Q) = 5000 J$
Electrical energy utilized by the motor
$=$ Work done by the motor $(W)$
$β = \frac{Q}{W}$
$∴ W = \frac{Q}{β} = \frac{5000}{5} = 1000 J$
$= 1 kJ$

AP EAMCET (22.09.2020) Shift-I

Thermodynamics

148541 The efficiency of a Carnot's engine is $100 %$ only when

1 ideal gas is used as a working substance

2 temperature of the sink is equal to absolute zero

3 source temperature is equal to the temperature of the sink

4 source temperature is equal to absolute zero

Explanation:

B The efficiency of Carnot's engine is given as
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $T_{1} =$ temperature of source
$T_{2} =$ temperature of sink
For efficiency $100 %, η = 1$
$1 = 1 - \frac{T_{2}}{T_{1}} \Rightarrow \frac{T_{2}}{T_{1}} = 0$
$∴ T_{2} = 0$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148542 An ideal heat engine has an efficiency $η$ . The coefficient of performance of the engine when driven backward will be

1

1 - (\frac{1}{η})

2

η - (\frac{1}{η})

3

(\frac{1}{η}) - 1

4

\frac{1}{1 - η}

Explanation:

C Ideal heat engine, efficiency is given as-
$η = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}} = 1 - η$
$\frac{T_{1}}{T_{2}} = \frac{1}{1 - η}$
When heat engine operated in backward direction, then coefficient of performance given as-
$α = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}} = \frac{T_{2}}{T_{1} - T_{2}} = \frac{1}{\frac{T_{1}}{T_{2}} - 1}$
From equation (i) and (ii),
$α = \frac{1}{\frac{1}{1 - η} - 1} = \frac{1}{\frac{1}{1 - η} - 1}$
$\frac{1 - η}{1 - 1 + η} = \frac{1 - η}{η} = (\frac{1}{η}) - 1$

AP EAMCET (21.09.2020) Shift-I

Thermodynamics

148543 When the temperature difference between the source and sink increases, the efficiency of the heat engine

1 Decreases

2 Increases

3 Is not affected

4 May increases or decrease

Explanation:

B The efficiency ( $η$ ) of the Carnot engine is given by-
$η = 1 - \frac{T_{L}}{T_{H}}$
Where,
$T_{L} = Temperature of the sink$
$T_{H} = Temperature of source$
- When the difference between the source and sink increases, it's means source temperature increases while sink temperature decreases.
- If the temperature of the source $(T_{H})$ is increases then the efficiency of the Carnot engine increases.

AP EAMCET (18.09.2020) Shift-II

Thermodynamics

148540 An electrical refrigerator with $β = 5$ extracts $5000 J$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.

1

1 kJ

2

0.5 kJ

3

0.8 kJ

4

1.2 kJ

Explanation:

A Given,
Coefficient of performance of refrigerator-
$β = 5$
Amount of heat removed $(Q) = 5000 J$
Electrical energy utilized by the motor
$=$ Work done by the motor $(W)$
$β = \frac{Q}{W}$
$∴ W = \frac{Q}{β} = \frac{5000}{5} = 1000 J$
$= 1 kJ$

AP EAMCET (22.09.2020) Shift-I

Thermodynamics

148541 The efficiency of a Carnot's engine is $100 %$ only when

1 ideal gas is used as a working substance

2 temperature of the sink is equal to absolute zero

3 source temperature is equal to the temperature of the sink

4 source temperature is equal to absolute zero

Explanation:

B The efficiency of Carnot's engine is given as
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $T_{1} =$ temperature of source
$T_{2} =$ temperature of sink
For efficiency $100 %, η = 1$
$1 = 1 - \frac{T_{2}}{T_{1}} \Rightarrow \frac{T_{2}}{T_{1}} = 0$
$∴ T_{2} = 0$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148542 An ideal heat engine has an efficiency $η$ . The coefficient of performance of the engine when driven backward will be

1

1 - (\frac{1}{η})

2

η - (\frac{1}{η})

3

(\frac{1}{η}) - 1

4

\frac{1}{1 - η}

Explanation:

C Ideal heat engine, efficiency is given as-
$η = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}} = 1 - η$
$\frac{T_{1}}{T_{2}} = \frac{1}{1 - η}$
When heat engine operated in backward direction, then coefficient of performance given as-
$α = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}} = \frac{T_{2}}{T_{1} - T_{2}} = \frac{1}{\frac{T_{1}}{T_{2}} - 1}$
From equation (i) and (ii),
$α = \frac{1}{\frac{1}{1 - η} - 1} = \frac{1}{\frac{1}{1 - η} - 1}$
$\frac{1 - η}{1 - 1 + η} = \frac{1 - η}{η} = (\frac{1}{η}) - 1$

AP EAMCET (21.09.2020) Shift-I

Thermodynamics

148543 When the temperature difference between the source and sink increases, the efficiency of the heat engine

1 Decreases

2 Increases

3 Is not affected

4 May increases or decrease

Explanation:

B The efficiency ( $η$ ) of the Carnot engine is given by-
$η = 1 - \frac{T_{L}}{T_{H}}$
Where,
$T_{L} = Temperature of the sink$
$T_{H} = Temperature of source$
- When the difference between the source and sink increases, it's means source temperature increases while sink temperature decreases.
- If the temperature of the source $(T_{H})$ is increases then the efficiency of the Carnot engine increases.

AP EAMCET (18.09.2020) Shift-II

Thermodynamics

148540 An electrical refrigerator with $β = 5$ extracts $5000 J$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.

1

1 kJ

2

0.5 kJ

3

0.8 kJ

4

1.2 kJ

Explanation:

A Given,
Coefficient of performance of refrigerator-
$β = 5$
Amount of heat removed $(Q) = 5000 J$
Electrical energy utilized by the motor
$=$ Work done by the motor $(W)$
$β = \frac{Q}{W}$
$∴ W = \frac{Q}{β} = \frac{5000}{5} = 1000 J$
$= 1 kJ$

AP EAMCET (22.09.2020) Shift-I

Thermodynamics

148541 The efficiency of a Carnot's engine is $100 %$ only when

1 ideal gas is used as a working substance

2 temperature of the sink is equal to absolute zero

3 source temperature is equal to the temperature of the sink

4 source temperature is equal to absolute zero

Explanation:

B The efficiency of Carnot's engine is given as
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $T_{1} =$ temperature of source
$T_{2} =$ temperature of sink
For efficiency $100 %, η = 1$
$1 = 1 - \frac{T_{2}}{T_{1}} \Rightarrow \frac{T_{2}}{T_{1}} = 0$
$∴ T_{2} = 0$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148542 An ideal heat engine has an efficiency $η$ . The coefficient of performance of the engine when driven backward will be

1

1 - (\frac{1}{η})

2

η - (\frac{1}{η})

3

(\frac{1}{η}) - 1

4

\frac{1}{1 - η}

Explanation:

C Ideal heat engine, efficiency is given as-
$η = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}} = 1 - η$
$\frac{T_{1}}{T_{2}} = \frac{1}{1 - η}$
When heat engine operated in backward direction, then coefficient of performance given as-
$α = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}} = \frac{T_{2}}{T_{1} - T_{2}} = \frac{1}{\frac{T_{1}}{T_{2}} - 1}$
From equation (i) and (ii),
$α = \frac{1}{\frac{1}{1 - η} - 1} = \frac{1}{\frac{1}{1 - η} - 1}$
$\frac{1 - η}{1 - 1 + η} = \frac{1 - η}{η} = (\frac{1}{η}) - 1$

AP EAMCET (21.09.2020) Shift-I

Thermodynamics

148543 When the temperature difference between the source and sink increases, the efficiency of the heat engine

1 Decreases

2 Increases

3 Is not affected

4 May increases or decrease

Explanation:

B The efficiency ( $η$ ) of the Carnot engine is given by-
$η = 1 - \frac{T_{L}}{T_{H}}$
Where,
$T_{L} = Temperature of the sink$
$T_{H} = Temperature of source$
- When the difference between the source and sink increases, it's means source temperature increases while sink temperature decreases.
- If the temperature of the source $(T_{H})$ is increases then the efficiency of the Carnot engine increases.

AP EAMCET (18.09.2020) Shift-II

Thermodynamics

148540 An electrical refrigerator with $β = 5$ extracts $5000 J$ from the contents of the refrigerator. During this process, find the electrical energy utilised by its motor.

1

1 kJ

2

0.5 kJ

3

0.8 kJ

4

1.2 kJ

Explanation:

A Given,
Coefficient of performance of refrigerator-
$β = 5$
Amount of heat removed $(Q) = 5000 J$
Electrical energy utilized by the motor
$=$ Work done by the motor $(W)$
$β = \frac{Q}{W}$
$∴ W = \frac{Q}{β} = \frac{5000}{5} = 1000 J$
$= 1 kJ$

AP EAMCET (22.09.2020) Shift-I

Thermodynamics

148541 The efficiency of a Carnot's engine is $100 %$ only when

1 ideal gas is used as a working substance

2 temperature of the sink is equal to absolute zero

3 source temperature is equal to the temperature of the sink

4 source temperature is equal to absolute zero

Explanation:

B The efficiency of Carnot's engine is given as
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $T_{1} =$ temperature of source
$T_{2} =$ temperature of sink
For efficiency $100 %, η = 1$
$1 = 1 - \frac{T_{2}}{T_{1}} \Rightarrow \frac{T_{2}}{T_{1}} = 0$
$∴ T_{2} = 0$

AP EAMCET (21.09.2020) Shift-II

Thermodynamics

148542 An ideal heat engine has an efficiency $η$ . The coefficient of performance of the engine when driven backward will be

1

1 - (\frac{1}{η})

2

η - (\frac{1}{η})

3

(\frac{1}{η}) - 1

4

\frac{1}{1 - η}

Explanation:

C Ideal heat engine, efficiency is given as-
$η = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}} = 1 - η$
$\frac{T_{1}}{T_{2}} = \frac{1}{1 - η}$
When heat engine operated in backward direction, then coefficient of performance given as-
$α = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}} = \frac{T_{2}}{T_{1} - T_{2}} = \frac{1}{\frac{T_{1}}{T_{2}} - 1}$
From equation (i) and (ii),
$α = \frac{1}{\frac{1}{1 - η} - 1} = \frac{1}{\frac{1}{1 - η} - 1}$
$\frac{1 - η}{1 - 1 + η} = \frac{1 - η}{η} = (\frac{1}{η}) - 1$

AP EAMCET (21.09.2020) Shift-I

Thermodynamics

148543 When the temperature difference between the source and sink increases, the efficiency of the heat engine

1 Decreases

2 Increases

3 Is not affected

4 May increases or decrease

Explanation:

B The efficiency ( $η$ ) of the Carnot engine is given by-
$η = 1 - \frac{T_{L}}{T_{H}}$
Where,
$T_{L} = Temperature of the sink$
$T_{H} = Temperature of source$
- When the difference between the source and sink increases, it's means source temperature increases while sink temperature decreases.
- If the temperature of the source $(T_{H})$ is increases then the efficiency of the Carnot engine increases.

AP EAMCET (18.09.2020) Shift-II

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