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Thermodynamics

148544 An ideal Carnot's engine with an efficiency of $30 %$ operates between a source and a sink. If the temperature of the source is $500 K$ , that of the sink is

1

27^{\circ} C

2

57^{\circ} C

3

77^{\circ} C

4

107^{\circ} C

Explanation:

C Given,
Efficiency of Carnot engine $(η) = 30 % = 0.3$ Temperature of source $(T_{1}) = 500 K$ We know,
$η = 1 - \frac{T_{2}}{T_{1}}$
$0.3 = 1 - \frac{T_{2}}{500} \Rightarrow \frac{T_{2}}{500} = 1 - 0.3$
$T_{2} = 0.7 \times 500 = 350 K$
$= (350 - 273)^{\circ} C$
$= 77^{\circ} C$

AP EAMCET (17.09.2020) Shift-II

Thermodynamics

148545 Two heat engines $X$ and $Y$ of same efficiency are connected in series in such a way that the sink of $X$ works as source of $Y$ . $X$ receives heat at $900 K$ and rejects some heat to its sink at TK and in turn $Y$ rejects heat to its sink at $400 K$ , then the temperature $T$ is

1

550 K

2

600 K

3

650 K

4

700 K

Explanation:

B According to the question a two heat engines are shown in the figure are connected in series combination,

Given,
Temperature of source of $X$ heat engine $(T_{1}) = 900 K$
Temperature of sink of $Y$ heat engine $(T_{2}) = 400 K$
Let, The Temperature of sink of $X$ which act as source of $Y$ is $T_{2}$
So, efficiency of $X$ heat engine,
$η_{X} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{900}$
Similarly, $η_{Y} = 1 - \frac{T_{3}}{T_{2}} = 1 - \frac{400}{T_{2}}$
$∵ η_{X} = η_{Y}$
$1 - \frac{T_{2}}{900} = 1 - \frac{400}{T_{2}}$
$T_{2}^{2} = 900 \times 400$
$T_{2} = 600 K$

AP EAMCET (23.04.2019) Shift-I

Thermodynamics

148546 A Carnot engine of efficiency $40 %$ , takes heat from a source maintained at a temperature of $500 K$ . It is desired to have an engine of efficiency $60 %$ . Then, the source temperature for the same sink temperature must be

1

650 K

2

750 K

3

550 K

4

850 K

Explanation:

B We know that,
Efficiency of Carnot engine,
$η = (1 - \frac{T_{2}}{T_{1}}) \times 100$
Where, $T_{2} =$ sink temperature
$T_{1} =$ source temperature
For $40 %$ efficiency, $T_{1} = 500 K$
$\frac{40}{100} = (1 - \frac{T_{2}}{500})$
$0.4 = 1 - \frac{T_{2}}{500}$
$0.6 = \frac{T_{2}}{500}$
$T_{2} = 300 K$
For $60 %$ efficiency the source temperature for the same sink temperature,
$\frac{60}{100} = 1 - \frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}} = 1 - \frac{6}{10}$
$\frac{T_{2}}{T_{1}} = \frac{4}{10} = 0.4$
$\frac{10}{4} \times T_{2} = T_{1}$
$T_{1} = \frac{10}{4} \times 300 [∵ T_{2} = 300 K]$
$T_{1} = 750 K$

AP EAMCET (21.04.2019) Shift-II

Thermodynamics

148547 Match the temperature of the source and sink ( $T_{1}$ and $T_{2}$ respectively) of a Carnot heat engine given in List-I with the corresponding efficiencies given in List - II.
| List-I | List-II |
| :--- | :--- |
| A. $T_{1} = 500 K, T_{2} = 300 K$ | (i) 0.2 |
| B. $T_{1} = 500 K, T_{2} = 350 K$ | (ii) 0.3 |
| C. $T_{1} = 800 K, T_{2} = 400 K$ | (iii) 0.4 |
| D. $T_{1} = 450 K, T_{2} = 360 K$ | (iv) 0.5 |
The correct match is
Codes

1 (iii) (A) (iv) (B) (ii) (C) (i) (D)

2 (iv) (A) (iii) (B) (ii) (C) (i) (D)

3 (iii) (A) (i) (B) (iv) (C) (ii) (D)

4 (iii) (A) (ii) (B) (iv) (C) (i) (D)

Explanation:

D We know, Efficiency of Carnot engine is,
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $T_{2} =$ Sink temperature
$T_{1} =$ Source temperature
Now, check the efficiencies:
(a) When $T_{1} = 500 K, T_{2} = 300 K η = 1 - \frac{300}{500} = 0.4$
(b) When $T_{1} = 500 K, T_{2} = 350 K η = 1 - \frac{350}{500} = 0.3$
(c) When $T_{1} = 800 K, T_{2} = 400 K η = 1 - \frac{400}{800} = 0.5$
(d) When $T_{1} = 450 K, T_{2} = 360 K, η = 1 - \frac{360}{450} = 0.2$

AP EAMCET (21.04.2019) Shift-I

Thermodynamics

148544 An ideal Carnot's engine with an efficiency of $30 %$ operates between a source and a sink. If the temperature of the source is $500 K$ , that of the sink is

1

27^{\circ} C

2

57^{\circ} C

3

77^{\circ} C

4

107^{\circ} C

Explanation:

C Given,
Efficiency of Carnot engine $(η) = 30 % = 0.3$ Temperature of source $(T_{1}) = 500 K$ We know,
$η = 1 - \frac{T_{2}}{T_{1}}$
$0.3 = 1 - \frac{T_{2}}{500} \Rightarrow \frac{T_{2}}{500} = 1 - 0.3$
$T_{2} = 0.7 \times 500 = 350 K$
$= (350 - 273)^{\circ} C$
$= 77^{\circ} C$

AP EAMCET (17.09.2020) Shift-II

Thermodynamics

148545 Two heat engines $X$ and $Y$ of same efficiency are connected in series in such a way that the sink of $X$ works as source of $Y$ . $X$ receives heat at $900 K$ and rejects some heat to its sink at TK and in turn $Y$ rejects heat to its sink at $400 K$ , then the temperature $T$ is

1

550 K

2

600 K

3

650 K

4

700 K

Explanation:

B According to the question a two heat engines are shown in the figure are connected in series combination,

Given,
Temperature of source of $X$ heat engine $(T_{1}) = 900 K$
Temperature of sink of $Y$ heat engine $(T_{2}) = 400 K$
Let, The Temperature of sink of $X$ which act as source of $Y$ is $T_{2}$
So, efficiency of $X$ heat engine,
$η_{X} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{900}$
Similarly, $η_{Y} = 1 - \frac{T_{3}}{T_{2}} = 1 - \frac{400}{T_{2}}$
$∵ η_{X} = η_{Y}$
$1 - \frac{T_{2}}{900} = 1 - \frac{400}{T_{2}}$
$T_{2}^{2} = 900 \times 400$
$T_{2} = 600 K$

AP EAMCET (23.04.2019) Shift-I

Thermodynamics

148546 A Carnot engine of efficiency $40 %$ , takes heat from a source maintained at a temperature of $500 K$ . It is desired to have an engine of efficiency $60 %$ . Then, the source temperature for the same sink temperature must be

1

650 K

2

750 K

3

550 K

4

850 K

Explanation:

B We know that,
Efficiency of Carnot engine,
$η = (1 - \frac{T_{2}}{T_{1}}) \times 100$
Where, $T_{2} =$ sink temperature
$T_{1} =$ source temperature
For $40 %$ efficiency, $T_{1} = 500 K$
$\frac{40}{100} = (1 - \frac{T_{2}}{500})$
$0.4 = 1 - \frac{T_{2}}{500}$
$0.6 = \frac{T_{2}}{500}$
$T_{2} = 300 K$
For $60 %$ efficiency the source temperature for the same sink temperature,
$\frac{60}{100} = 1 - \frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}} = 1 - \frac{6}{10}$
$\frac{T_{2}}{T_{1}} = \frac{4}{10} = 0.4$
$\frac{10}{4} \times T_{2} = T_{1}$
$T_{1} = \frac{10}{4} \times 300 [∵ T_{2} = 300 K]$
$T_{1} = 750 K$

AP EAMCET (21.04.2019) Shift-II

Thermodynamics

148547 Match the temperature of the source and sink ( $T_{1}$ and $T_{2}$ respectively) of a Carnot heat engine given in List-I with the corresponding efficiencies given in List - II.
| List-I | List-II |
| :--- | :--- |
| A. $T_{1} = 500 K, T_{2} = 300 K$ | (i) 0.2 |
| B. $T_{1} = 500 K, T_{2} = 350 K$ | (ii) 0.3 |
| C. $T_{1} = 800 K, T_{2} = 400 K$ | (iii) 0.4 |
| D. $T_{1} = 450 K, T_{2} = 360 K$ | (iv) 0.5 |
The correct match is
Codes

1 (iii) (A) (iv) (B) (ii) (C) (i) (D)

2 (iv) (A) (iii) (B) (ii) (C) (i) (D)

3 (iii) (A) (i) (B) (iv) (C) (ii) (D)

4 (iii) (A) (ii) (B) (iv) (C) (i) (D)

Explanation:

D We know, Efficiency of Carnot engine is,
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $T_{2} =$ Sink temperature
$T_{1} =$ Source temperature
Now, check the efficiencies:
(a) When $T_{1} = 500 K, T_{2} = 300 K η = 1 - \frac{300}{500} = 0.4$
(b) When $T_{1} = 500 K, T_{2} = 350 K η = 1 - \frac{350}{500} = 0.3$
(c) When $T_{1} = 800 K, T_{2} = 400 K η = 1 - \frac{400}{800} = 0.5$
(d) When $T_{1} = 450 K, T_{2} = 360 K, η = 1 - \frac{360}{450} = 0.2$

AP EAMCET (21.04.2019) Shift-I

Thermodynamics

148544 An ideal Carnot's engine with an efficiency of $30 %$ operates between a source and a sink. If the temperature of the source is $500 K$ , that of the sink is

1

27^{\circ} C

2

57^{\circ} C

3

77^{\circ} C

4

107^{\circ} C

Explanation:

C Given,
Efficiency of Carnot engine $(η) = 30 % = 0.3$ Temperature of source $(T_{1}) = 500 K$ We know,
$η = 1 - \frac{T_{2}}{T_{1}}$
$0.3 = 1 - \frac{T_{2}}{500} \Rightarrow \frac{T_{2}}{500} = 1 - 0.3$
$T_{2} = 0.7 \times 500 = 350 K$
$= (350 - 273)^{\circ} C$
$= 77^{\circ} C$

AP EAMCET (17.09.2020) Shift-II

Thermodynamics

148545 Two heat engines $X$ and $Y$ of same efficiency are connected in series in such a way that the sink of $X$ works as source of $Y$ . $X$ receives heat at $900 K$ and rejects some heat to its sink at TK and in turn $Y$ rejects heat to its sink at $400 K$ , then the temperature $T$ is

1

550 K

2

600 K

3

650 K

4

700 K

Explanation:

B According to the question a two heat engines are shown in the figure are connected in series combination,

Given,
Temperature of source of $X$ heat engine $(T_{1}) = 900 K$
Temperature of sink of $Y$ heat engine $(T_{2}) = 400 K$
Let, The Temperature of sink of $X$ which act as source of $Y$ is $T_{2}$
So, efficiency of $X$ heat engine,
$η_{X} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{900}$
Similarly, $η_{Y} = 1 - \frac{T_{3}}{T_{2}} = 1 - \frac{400}{T_{2}}$
$∵ η_{X} = η_{Y}$
$1 - \frac{T_{2}}{900} = 1 - \frac{400}{T_{2}}$
$T_{2}^{2} = 900 \times 400$
$T_{2} = 600 K$

AP EAMCET (23.04.2019) Shift-I

Thermodynamics

148546 A Carnot engine of efficiency $40 %$ , takes heat from a source maintained at a temperature of $500 K$ . It is desired to have an engine of efficiency $60 %$ . Then, the source temperature for the same sink temperature must be

1

650 K

2

750 K

3

550 K

4

850 K

Explanation:

B We know that,
Efficiency of Carnot engine,
$η = (1 - \frac{T_{2}}{T_{1}}) \times 100$
Where, $T_{2} =$ sink temperature
$T_{1} =$ source temperature
For $40 %$ efficiency, $T_{1} = 500 K$
$\frac{40}{100} = (1 - \frac{T_{2}}{500})$
$0.4 = 1 - \frac{T_{2}}{500}$
$0.6 = \frac{T_{2}}{500}$
$T_{2} = 300 K$
For $60 %$ efficiency the source temperature for the same sink temperature,
$\frac{60}{100} = 1 - \frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}} = 1 - \frac{6}{10}$
$\frac{T_{2}}{T_{1}} = \frac{4}{10} = 0.4$
$\frac{10}{4} \times T_{2} = T_{1}$
$T_{1} = \frac{10}{4} \times 300 [∵ T_{2} = 300 K]$
$T_{1} = 750 K$

AP EAMCET (21.04.2019) Shift-II

Thermodynamics

148547 Match the temperature of the source and sink ( $T_{1}$ and $T_{2}$ respectively) of a Carnot heat engine given in List-I with the corresponding efficiencies given in List - II.
| List-I | List-II |
| :--- | :--- |
| A. $T_{1} = 500 K, T_{2} = 300 K$ | (i) 0.2 |
| B. $T_{1} = 500 K, T_{2} = 350 K$ | (ii) 0.3 |
| C. $T_{1} = 800 K, T_{2} = 400 K$ | (iii) 0.4 |
| D. $T_{1} = 450 K, T_{2} = 360 K$ | (iv) 0.5 |
The correct match is
Codes

1 (iii) (A) (iv) (B) (ii) (C) (i) (D)

2 (iv) (A) (iii) (B) (ii) (C) (i) (D)

3 (iii) (A) (i) (B) (iv) (C) (ii) (D)

4 (iii) (A) (ii) (B) (iv) (C) (i) (D)

Explanation:

D We know, Efficiency of Carnot engine is,
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $T_{2} =$ Sink temperature
$T_{1} =$ Source temperature
Now, check the efficiencies:
(a) When $T_{1} = 500 K, T_{2} = 300 K η = 1 - \frac{300}{500} = 0.4$
(b) When $T_{1} = 500 K, T_{2} = 350 K η = 1 - \frac{350}{500} = 0.3$
(c) When $T_{1} = 800 K, T_{2} = 400 K η = 1 - \frac{400}{800} = 0.5$
(d) When $T_{1} = 450 K, T_{2} = 360 K, η = 1 - \frac{360}{450} = 0.2$

AP EAMCET (21.04.2019) Shift-I

Thermodynamics

148544 An ideal Carnot's engine with an efficiency of $30 %$ operates between a source and a sink. If the temperature of the source is $500 K$ , that of the sink is

1

27^{\circ} C

2

57^{\circ} C

3

77^{\circ} C

4

107^{\circ} C

Explanation:

C Given,
Efficiency of Carnot engine $(η) = 30 % = 0.3$ Temperature of source $(T_{1}) = 500 K$ We know,
$η = 1 - \frac{T_{2}}{T_{1}}$
$0.3 = 1 - \frac{T_{2}}{500} \Rightarrow \frac{T_{2}}{500} = 1 - 0.3$
$T_{2} = 0.7 \times 500 = 350 K$
$= (350 - 273)^{\circ} C$
$= 77^{\circ} C$

AP EAMCET (17.09.2020) Shift-II

Thermodynamics

148545 Two heat engines $X$ and $Y$ of same efficiency are connected in series in such a way that the sink of $X$ works as source of $Y$ . $X$ receives heat at $900 K$ and rejects some heat to its sink at TK and in turn $Y$ rejects heat to its sink at $400 K$ , then the temperature $T$ is

1

550 K

2

600 K

3

650 K

4

700 K

Explanation:

B According to the question a two heat engines are shown in the figure are connected in series combination,

Given,
Temperature of source of $X$ heat engine $(T_{1}) = 900 K$
Temperature of sink of $Y$ heat engine $(T_{2}) = 400 K$
Let, The Temperature of sink of $X$ which act as source of $Y$ is $T_{2}$
So, efficiency of $X$ heat engine,
$η_{X} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{900}$
Similarly, $η_{Y} = 1 - \frac{T_{3}}{T_{2}} = 1 - \frac{400}{T_{2}}$
$∵ η_{X} = η_{Y}$
$1 - \frac{T_{2}}{900} = 1 - \frac{400}{T_{2}}$
$T_{2}^{2} = 900 \times 400$
$T_{2} = 600 K$

AP EAMCET (23.04.2019) Shift-I

Thermodynamics

148546 A Carnot engine of efficiency $40 %$ , takes heat from a source maintained at a temperature of $500 K$ . It is desired to have an engine of efficiency $60 %$ . Then, the source temperature for the same sink temperature must be

1

650 K

2

750 K

3

550 K

4

850 K

Explanation:

B We know that,
Efficiency of Carnot engine,
$η = (1 - \frac{T_{2}}{T_{1}}) \times 100$
Where, $T_{2} =$ sink temperature
$T_{1} =$ source temperature
For $40 %$ efficiency, $T_{1} = 500 K$
$\frac{40}{100} = (1 - \frac{T_{2}}{500})$
$0.4 = 1 - \frac{T_{2}}{500}$
$0.6 = \frac{T_{2}}{500}$
$T_{2} = 300 K$
For $60 %$ efficiency the source temperature for the same sink temperature,
$\frac{60}{100} = 1 - \frac{T_{2}}{T_{1}}$
$\frac{T_{2}}{T_{1}} = 1 - \frac{6}{10}$
$\frac{T_{2}}{T_{1}} = \frac{4}{10} = 0.4$
$\frac{10}{4} \times T_{2} = T_{1}$
$T_{1} = \frac{10}{4} \times 300 [∵ T_{2} = 300 K]$
$T_{1} = 750 K$

AP EAMCET (21.04.2019) Shift-II

Thermodynamics

148547 Match the temperature of the source and sink ( $T_{1}$ and $T_{2}$ respectively) of a Carnot heat engine given in List-I with the corresponding efficiencies given in List - II.
| List-I | List-II |
| :--- | :--- |
| A. $T_{1} = 500 K, T_{2} = 300 K$ | (i) 0.2 |
| B. $T_{1} = 500 K, T_{2} = 350 K$ | (ii) 0.3 |
| C. $T_{1} = 800 K, T_{2} = 400 K$ | (iii) 0.4 |
| D. $T_{1} = 450 K, T_{2} = 360 K$ | (iv) 0.5 |
The correct match is
Codes

1 (iii) (A) (iv) (B) (ii) (C) (i) (D)

2 (iv) (A) (iii) (B) (ii) (C) (i) (D)

3 (iii) (A) (i) (B) (iv) (C) (ii) (D)

4 (iii) (A) (ii) (B) (iv) (C) (i) (D)

Explanation:

D We know, Efficiency of Carnot engine is,
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $T_{2} =$ Sink temperature
$T_{1} =$ Source temperature
Now, check the efficiencies:
(a) When $T_{1} = 500 K, T_{2} = 300 K η = 1 - \frac{300}{500} = 0.4$
(b) When $T_{1} = 500 K, T_{2} = 350 K η = 1 - \frac{350}{500} = 0.3$
(c) When $T_{1} = 800 K, T_{2} = 400 K η = 1 - \frac{400}{800} = 0.5$
(d) When $T_{1} = 450 K, T_{2} = 360 K, η = 1 - \frac{360}{450} = 0.2$

AP EAMCET (21.04.2019) Shift-I

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