QBManager

Thermodynamics

148518 A Carnot engine with efficiency $50 %$ takes heat from a source at $600 K$ , In order to increase the efficiency to $70 %$ , keeping the temperature of sink same, the new temperature of the source will be.

1

900 K

2

300 K

3

1000 K

4

360 K

Explanation:

C $η = 1 - \frac{T_{sink}}{T_{source}}$
$\frac{50}{100} = 1 - \frac{T_{sink}}{600}$
$\frac{T_{sink}}{600} = \frac{1}{2}$
$\Rightarrow T_{sink} = 300$
To increase in efficiency by $70 %$ ,
$\frac{70}{100} = 1 - \frac{300}{T_{source}^{'}}$
$\frac{300}{T_{source}^{'}} = 1 - \frac{70}{100}$
$\frac{300}{T_{source}^{'}} = \frac{3}{10}$
$T_{source}^{'} = 1000 K$

JEE Main-25.01.2023

Thermodynamics

148519 A Carnot engine operating between two reservoirs has efficiency $1 / 3$ . When the temperature of cold reservoir raised by $x$ , its efficiency decreases to $1 / 6$ . The value of $x$ , if the temperature of hot reservoir is $99^{\circ} C$ , will be

1

62 K

2

16.5 K

3

33 K

4

66 K

Explanation:

A
Let $T_{2}$ is the temperature of sink and $T_{1}$ is the temperature of source,
Given, $T_{1} = 99^{\circ} C + 273 = 372 K$
$η_{1} = \frac{1}{3}, η_{2} = \frac{1}{6}$
$η_{1} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{372}$
$\frac{1}{3} = 1 - \frac{T_{2}}{372}$
$\frac{T_{2}}{372} = 1 - \frac{1}{3} = \frac{2}{3}$
$T_{2} = 248 K (sink temperature)$
Now the sink temperature $(T_{2} = 248 K)$ is increased by $x$ $&$ efficiency become $\frac{1}{6}$ ,
$η_{2} = 1 - \frac{T_{2} + x}{T_{1}}$
$\frac{1}{6} = 1 - \frac{248 + x}{372}$
$\frac{248 + x}{372} = 1 - \frac{1}{6} = \frac{5}{6}$
$248 + x = \frac{5}{6} \times 372$
$248 + x = 5 \times 62$
$248 + x = 310$
$x = 62 K$

JEE Main-01.02.2023

Thermodynamics

148520 A Carnot engine whose low temperature reservoir is at $350 K$ has an efficiency of $50 %$ It is desired to increase this to $60 %$ . If the temperature of the low temperature reservoir remains constant, then the temperature of high temperature reservoir must be increased by how many degrees?

1 15

2 175

3 100

4 50

5 120

Explanation:

B Efficiency of Carnot engine-
$η = 1 - \frac{T_{cold}}{T_{hot}}$
When, $η = 50 %$
$\frac{50}{100} = 1 - \frac{T_{cold}}{T_{hot}}$
$0.5 = 1 - \frac{350}{T_{hot}} \Rightarrow T_{hot} = 700 K$
When, $η = 60 %$
$\frac{60}{100} = 1 - \frac{T_{cold}}{T_{hot}^{'}}$
$\frac{60}{100} = 1 - \frac{350}{T_{hot}^{'}}$
$T_{hot}^{'} = 875 K$
Hence the temperature of high reservoir is increased by-
$T_{hot}^{'} - T_{hot} = 875 K - 700 K$
$= 175 K or 175^{\circ} C$

Kerala CEE -2018

Thermodynamics

148522 A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency 0.2 . When $T_{2}$ is reduced by $50 K$ , its efficiency increases to 0.4 . Then, $T_{1}$ and $T_{2}$ are respectively

1

200 K, 150 K

2

250 K, 200 K

3

300 K, 250 K

4

300 K, 200 K

5

300 K, 150 K

Explanation:

B By Carnot's ideal heat engine-
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $η_{1} = 0.2 & η_{2} = 0.4$
For the first condition-
$η_{1} = 1 - \frac{T_{2}}{T_{1}} \Rightarrow 0.2 = 1 - \frac{T_{2}}{T_{1}}$
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
For the second condition-
$η_{2} = 1 - \frac{T_{2} - 50}{T_{1}}$
$0.4 = \frac{T_{1} - (T_{2} - 50)}{T_{1}}$
$0.4 = \frac{T_{1} - T_{2} + 50}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}} + \frac{50}{T_{1}}$
$0.4 = 0.2 + \frac{50}{T_{1}}$
$0.2 = \frac{50}{T_{1}}$
$T_{1} = 250 K$
Put the value of $T_{1}$ in equation (i), we get-
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
$T_{2} = T_{1} - 0.2 T_{1}$
$T_{2} = 250 - 0.2 \times 250 = 250 - 50$
$T_{2} = 200 K$

Kerala CEE- 2014

Thermodynamics

148524 A Carnot engine whose efficiency is $40 %$ , receives heat at $500 K$ . If the efficiency is to be $50 %$ , the source temperature for the same exhaust temperature is

1

900 K

2

600 K

3

700 K

4

800 K

5

550 K

Explanation:

B Carnot efficiency relation-
$η = \frac{T_{1} - T_{2}}{T_{1}}$
Case-1 $\frac{T_{1} - T_{2}}{T_{1}} = 0.4$
$T_{1} - T_{2} = 0.4 T_{1}$
$T_{2} = 0.6 T_{1}$
Case-2 $\frac{T_{1}^{'} - T_{2}}{T_{1}^{'}} = 0.5$
$T_{1}^{'} = \frac{0.6}{0.5} T_{1} = \frac{6}{5} \times 500$
$T_{1}^{'} = 600 K$

Kerala CEE - 2011

Thermodynamics

148518 A Carnot engine with efficiency $50 %$ takes heat from a source at $600 K$ , In order to increase the efficiency to $70 %$ , keeping the temperature of sink same, the new temperature of the source will be.

1

900 K

2

300 K

3

1000 K

4

360 K

Explanation:

C $η = 1 - \frac{T_{sink}}{T_{source}}$
$\frac{50}{100} = 1 - \frac{T_{sink}}{600}$
$\frac{T_{sink}}{600} = \frac{1}{2}$
$\Rightarrow T_{sink} = 300$
To increase in efficiency by $70 %$ ,
$\frac{70}{100} = 1 - \frac{300}{T_{source}^{'}}$
$\frac{300}{T_{source}^{'}} = 1 - \frac{70}{100}$
$\frac{300}{T_{source}^{'}} = \frac{3}{10}$
$T_{source}^{'} = 1000 K$

JEE Main-25.01.2023

Thermodynamics

148519 A Carnot engine operating between two reservoirs has efficiency $1 / 3$ . When the temperature of cold reservoir raised by $x$ , its efficiency decreases to $1 / 6$ . The value of $x$ , if the temperature of hot reservoir is $99^{\circ} C$ , will be

1

62 K

2

16.5 K

3

33 K

4

66 K

Explanation:

A
Let $T_{2}$ is the temperature of sink and $T_{1}$ is the temperature of source,
Given, $T_{1} = 99^{\circ} C + 273 = 372 K$
$η_{1} = \frac{1}{3}, η_{2} = \frac{1}{6}$
$η_{1} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{372}$
$\frac{1}{3} = 1 - \frac{T_{2}}{372}$
$\frac{T_{2}}{372} = 1 - \frac{1}{3} = \frac{2}{3}$
$T_{2} = 248 K (sink temperature)$
Now the sink temperature $(T_{2} = 248 K)$ is increased by $x$ $&$ efficiency become $\frac{1}{6}$ ,
$η_{2} = 1 - \frac{T_{2} + x}{T_{1}}$
$\frac{1}{6} = 1 - \frac{248 + x}{372}$
$\frac{248 + x}{372} = 1 - \frac{1}{6} = \frac{5}{6}$
$248 + x = \frac{5}{6} \times 372$
$248 + x = 5 \times 62$
$248 + x = 310$
$x = 62 K$

JEE Main-01.02.2023

Thermodynamics

148520 A Carnot engine whose low temperature reservoir is at $350 K$ has an efficiency of $50 %$ It is desired to increase this to $60 %$ . If the temperature of the low temperature reservoir remains constant, then the temperature of high temperature reservoir must be increased by how many degrees?

1 15

2 175

3 100

4 50

5 120

Explanation:

B Efficiency of Carnot engine-
$η = 1 - \frac{T_{cold}}{T_{hot}}$
When, $η = 50 %$
$\frac{50}{100} = 1 - \frac{T_{cold}}{T_{hot}}$
$0.5 = 1 - \frac{350}{T_{hot}} \Rightarrow T_{hot} = 700 K$
When, $η = 60 %$
$\frac{60}{100} = 1 - \frac{T_{cold}}{T_{hot}^{'}}$
$\frac{60}{100} = 1 - \frac{350}{T_{hot}^{'}}$
$T_{hot}^{'} = 875 K$
Hence the temperature of high reservoir is increased by-
$T_{hot}^{'} - T_{hot} = 875 K - 700 K$
$= 175 K or 175^{\circ} C$

Kerala CEE -2018

Thermodynamics

148522 A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency 0.2 . When $T_{2}$ is reduced by $50 K$ , its efficiency increases to 0.4 . Then, $T_{1}$ and $T_{2}$ are respectively

1

200 K, 150 K

2

250 K, 200 K

3

300 K, 250 K

4

300 K, 200 K

5

300 K, 150 K

Explanation:

B By Carnot's ideal heat engine-
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $η_{1} = 0.2 & η_{2} = 0.4$
For the first condition-
$η_{1} = 1 - \frac{T_{2}}{T_{1}} \Rightarrow 0.2 = 1 - \frac{T_{2}}{T_{1}}$
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
For the second condition-
$η_{2} = 1 - \frac{T_{2} - 50}{T_{1}}$
$0.4 = \frac{T_{1} - (T_{2} - 50)}{T_{1}}$
$0.4 = \frac{T_{1} - T_{2} + 50}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}} + \frac{50}{T_{1}}$
$0.4 = 0.2 + \frac{50}{T_{1}}$
$0.2 = \frac{50}{T_{1}}$
$T_{1} = 250 K$
Put the value of $T_{1}$ in equation (i), we get-
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
$T_{2} = T_{1} - 0.2 T_{1}$
$T_{2} = 250 - 0.2 \times 250 = 250 - 50$
$T_{2} = 200 K$

Kerala CEE- 2014

Thermodynamics

148524 A Carnot engine whose efficiency is $40 %$ , receives heat at $500 K$ . If the efficiency is to be $50 %$ , the source temperature for the same exhaust temperature is

1

900 K

2

600 K

3

700 K

4

800 K

5

550 K

Explanation:

B Carnot efficiency relation-
$η = \frac{T_{1} - T_{2}}{T_{1}}$
Case-1 $\frac{T_{1} - T_{2}}{T_{1}} = 0.4$
$T_{1} - T_{2} = 0.4 T_{1}$
$T_{2} = 0.6 T_{1}$
Case-2 $\frac{T_{1}^{'} - T_{2}}{T_{1}^{'}} = 0.5$
$T_{1}^{'} = \frac{0.6}{0.5} T_{1} = \frac{6}{5} \times 500$
$T_{1}^{'} = 600 K$

Kerala CEE - 2011

Thermodynamics

148518 A Carnot engine with efficiency $50 %$ takes heat from a source at $600 K$ , In order to increase the efficiency to $70 %$ , keeping the temperature of sink same, the new temperature of the source will be.

1

900 K

2

300 K

3

1000 K

4

360 K

Explanation:

C $η = 1 - \frac{T_{sink}}{T_{source}}$
$\frac{50}{100} = 1 - \frac{T_{sink}}{600}$
$\frac{T_{sink}}{600} = \frac{1}{2}$
$\Rightarrow T_{sink} = 300$
To increase in efficiency by $70 %$ ,
$\frac{70}{100} = 1 - \frac{300}{T_{source}^{'}}$
$\frac{300}{T_{source}^{'}} = 1 - \frac{70}{100}$
$\frac{300}{T_{source}^{'}} = \frac{3}{10}$
$T_{source}^{'} = 1000 K$

JEE Main-25.01.2023

Thermodynamics

148519 A Carnot engine operating between two reservoirs has efficiency $1 / 3$ . When the temperature of cold reservoir raised by $x$ , its efficiency decreases to $1 / 6$ . The value of $x$ , if the temperature of hot reservoir is $99^{\circ} C$ , will be

1

62 K

2

16.5 K

3

33 K

4

66 K

Explanation:

A
Let $T_{2}$ is the temperature of sink and $T_{1}$ is the temperature of source,
Given, $T_{1} = 99^{\circ} C + 273 = 372 K$
$η_{1} = \frac{1}{3}, η_{2} = \frac{1}{6}$
$η_{1} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{372}$
$\frac{1}{3} = 1 - \frac{T_{2}}{372}$
$\frac{T_{2}}{372} = 1 - \frac{1}{3} = \frac{2}{3}$
$T_{2} = 248 K (sink temperature)$
Now the sink temperature $(T_{2} = 248 K)$ is increased by $x$ $&$ efficiency become $\frac{1}{6}$ ,
$η_{2} = 1 - \frac{T_{2} + x}{T_{1}}$
$\frac{1}{6} = 1 - \frac{248 + x}{372}$
$\frac{248 + x}{372} = 1 - \frac{1}{6} = \frac{5}{6}$
$248 + x = \frac{5}{6} \times 372$
$248 + x = 5 \times 62$
$248 + x = 310$
$x = 62 K$

JEE Main-01.02.2023

Thermodynamics

148520 A Carnot engine whose low temperature reservoir is at $350 K$ has an efficiency of $50 %$ It is desired to increase this to $60 %$ . If the temperature of the low temperature reservoir remains constant, then the temperature of high temperature reservoir must be increased by how many degrees?

1 15

2 175

3 100

4 50

5 120

Explanation:

B Efficiency of Carnot engine-
$η = 1 - \frac{T_{cold}}{T_{hot}}$
When, $η = 50 %$
$\frac{50}{100} = 1 - \frac{T_{cold}}{T_{hot}}$
$0.5 = 1 - \frac{350}{T_{hot}} \Rightarrow T_{hot} = 700 K$
When, $η = 60 %$
$\frac{60}{100} = 1 - \frac{T_{cold}}{T_{hot}^{'}}$
$\frac{60}{100} = 1 - \frac{350}{T_{hot}^{'}}$
$T_{hot}^{'} = 875 K$
Hence the temperature of high reservoir is increased by-
$T_{hot}^{'} - T_{hot} = 875 K - 700 K$
$= 175 K or 175^{\circ} C$

Kerala CEE -2018

Thermodynamics

148522 A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency 0.2 . When $T_{2}$ is reduced by $50 K$ , its efficiency increases to 0.4 . Then, $T_{1}$ and $T_{2}$ are respectively

1

200 K, 150 K

2

250 K, 200 K

3

300 K, 250 K

4

300 K, 200 K

5

300 K, 150 K

Explanation:

B By Carnot's ideal heat engine-
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $η_{1} = 0.2 & η_{2} = 0.4$
For the first condition-
$η_{1} = 1 - \frac{T_{2}}{T_{1}} \Rightarrow 0.2 = 1 - \frac{T_{2}}{T_{1}}$
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
For the second condition-
$η_{2} = 1 - \frac{T_{2} - 50}{T_{1}}$
$0.4 = \frac{T_{1} - (T_{2} - 50)}{T_{1}}$
$0.4 = \frac{T_{1} - T_{2} + 50}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}} + \frac{50}{T_{1}}$
$0.4 = 0.2 + \frac{50}{T_{1}}$
$0.2 = \frac{50}{T_{1}}$
$T_{1} = 250 K$
Put the value of $T_{1}$ in equation (i), we get-
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
$T_{2} = T_{1} - 0.2 T_{1}$
$T_{2} = 250 - 0.2 \times 250 = 250 - 50$
$T_{2} = 200 K$

Kerala CEE- 2014

Thermodynamics

148524 A Carnot engine whose efficiency is $40 %$ , receives heat at $500 K$ . If the efficiency is to be $50 %$ , the source temperature for the same exhaust temperature is

1

900 K

2

600 K

3

700 K

4

800 K

5

550 K

Explanation:

B Carnot efficiency relation-
$η = \frac{T_{1} - T_{2}}{T_{1}}$
Case-1 $\frac{T_{1} - T_{2}}{T_{1}} = 0.4$
$T_{1} - T_{2} = 0.4 T_{1}$
$T_{2} = 0.6 T_{1}$
Case-2 $\frac{T_{1}^{'} - T_{2}}{T_{1}^{'}} = 0.5$
$T_{1}^{'} = \frac{0.6}{0.5} T_{1} = \frac{6}{5} \times 500$
$T_{1}^{'} = 600 K$

Kerala CEE - 2011

Thermodynamics

148518 A Carnot engine with efficiency $50 %$ takes heat from a source at $600 K$ , In order to increase the efficiency to $70 %$ , keeping the temperature of sink same, the new temperature of the source will be.

1

900 K

2

300 K

3

1000 K

4

360 K

Explanation:

C $η = 1 - \frac{T_{sink}}{T_{source}}$
$\frac{50}{100} = 1 - \frac{T_{sink}}{600}$
$\frac{T_{sink}}{600} = \frac{1}{2}$
$\Rightarrow T_{sink} = 300$
To increase in efficiency by $70 %$ ,
$\frac{70}{100} = 1 - \frac{300}{T_{source}^{'}}$
$\frac{300}{T_{source}^{'}} = 1 - \frac{70}{100}$
$\frac{300}{T_{source}^{'}} = \frac{3}{10}$
$T_{source}^{'} = 1000 K$

JEE Main-25.01.2023

Thermodynamics

148519 A Carnot engine operating between two reservoirs has efficiency $1 / 3$ . When the temperature of cold reservoir raised by $x$ , its efficiency decreases to $1 / 6$ . The value of $x$ , if the temperature of hot reservoir is $99^{\circ} C$ , will be

1

62 K

2

16.5 K

3

33 K

4

66 K

Explanation:

A
Let $T_{2}$ is the temperature of sink and $T_{1}$ is the temperature of source,
Given, $T_{1} = 99^{\circ} C + 273 = 372 K$
$η_{1} = \frac{1}{3}, η_{2} = \frac{1}{6}$
$η_{1} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{372}$
$\frac{1}{3} = 1 - \frac{T_{2}}{372}$
$\frac{T_{2}}{372} = 1 - \frac{1}{3} = \frac{2}{3}$
$T_{2} = 248 K (sink temperature)$
Now the sink temperature $(T_{2} = 248 K)$ is increased by $x$ $&$ efficiency become $\frac{1}{6}$ ,
$η_{2} = 1 - \frac{T_{2} + x}{T_{1}}$
$\frac{1}{6} = 1 - \frac{248 + x}{372}$
$\frac{248 + x}{372} = 1 - \frac{1}{6} = \frac{5}{6}$
$248 + x = \frac{5}{6} \times 372$
$248 + x = 5 \times 62$
$248 + x = 310$
$x = 62 K$

JEE Main-01.02.2023

Thermodynamics

148520 A Carnot engine whose low temperature reservoir is at $350 K$ has an efficiency of $50 %$ It is desired to increase this to $60 %$ . If the temperature of the low temperature reservoir remains constant, then the temperature of high temperature reservoir must be increased by how many degrees?

1 15

2 175

3 100

4 50

5 120

Explanation:

B Efficiency of Carnot engine-
$η = 1 - \frac{T_{cold}}{T_{hot}}$
When, $η = 50 %$
$\frac{50}{100} = 1 - \frac{T_{cold}}{T_{hot}}$
$0.5 = 1 - \frac{350}{T_{hot}} \Rightarrow T_{hot} = 700 K$
When, $η = 60 %$
$\frac{60}{100} = 1 - \frac{T_{cold}}{T_{hot}^{'}}$
$\frac{60}{100} = 1 - \frac{350}{T_{hot}^{'}}$
$T_{hot}^{'} = 875 K$
Hence the temperature of high reservoir is increased by-
$T_{hot}^{'} - T_{hot} = 875 K - 700 K$
$= 175 K or 175^{\circ} C$

Kerala CEE -2018

Thermodynamics

148522 A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency 0.2 . When $T_{2}$ is reduced by $50 K$ , its efficiency increases to 0.4 . Then, $T_{1}$ and $T_{2}$ are respectively

1

200 K, 150 K

2

250 K, 200 K

3

300 K, 250 K

4

300 K, 200 K

5

300 K, 150 K

Explanation:

B By Carnot's ideal heat engine-
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $η_{1} = 0.2 & η_{2} = 0.4$
For the first condition-
$η_{1} = 1 - \frac{T_{2}}{T_{1}} \Rightarrow 0.2 = 1 - \frac{T_{2}}{T_{1}}$
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
For the second condition-
$η_{2} = 1 - \frac{T_{2} - 50}{T_{1}}$
$0.4 = \frac{T_{1} - (T_{2} - 50)}{T_{1}}$
$0.4 = \frac{T_{1} - T_{2} + 50}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}} + \frac{50}{T_{1}}$
$0.4 = 0.2 + \frac{50}{T_{1}}$
$0.2 = \frac{50}{T_{1}}$
$T_{1} = 250 K$
Put the value of $T_{1}$ in equation (i), we get-
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
$T_{2} = T_{1} - 0.2 T_{1}$
$T_{2} = 250 - 0.2 \times 250 = 250 - 50$
$T_{2} = 200 K$

Kerala CEE- 2014

Thermodynamics

148524 A Carnot engine whose efficiency is $40 %$ , receives heat at $500 K$ . If the efficiency is to be $50 %$ , the source temperature for the same exhaust temperature is

1

900 K

2

600 K

3

700 K

4

800 K

5

550 K

Explanation:

B Carnot efficiency relation-
$η = \frac{T_{1} - T_{2}}{T_{1}}$
Case-1 $\frac{T_{1} - T_{2}}{T_{1}} = 0.4$
$T_{1} - T_{2} = 0.4 T_{1}$
$T_{2} = 0.6 T_{1}$
Case-2 $\frac{T_{1}^{'} - T_{2}}{T_{1}^{'}} = 0.5$
$T_{1}^{'} = \frac{0.6}{0.5} T_{1} = \frac{6}{5} \times 500$
$T_{1}^{'} = 600 K$

Kerala CEE - 2011

Thermodynamics

148518 A Carnot engine with efficiency $50 %$ takes heat from a source at $600 K$ , In order to increase the efficiency to $70 %$ , keeping the temperature of sink same, the new temperature of the source will be.

1

900 K

2

300 K

3

1000 K

4

360 K

Explanation:

C $η = 1 - \frac{T_{sink}}{T_{source}}$
$\frac{50}{100} = 1 - \frac{T_{sink}}{600}$
$\frac{T_{sink}}{600} = \frac{1}{2}$
$\Rightarrow T_{sink} = 300$
To increase in efficiency by $70 %$ ,
$\frac{70}{100} = 1 - \frac{300}{T_{source}^{'}}$
$\frac{300}{T_{source}^{'}} = 1 - \frac{70}{100}$
$\frac{300}{T_{source}^{'}} = \frac{3}{10}$
$T_{source}^{'} = 1000 K$

JEE Main-25.01.2023

Thermodynamics

148519 A Carnot engine operating between two reservoirs has efficiency $1 / 3$ . When the temperature of cold reservoir raised by $x$ , its efficiency decreases to $1 / 6$ . The value of $x$ , if the temperature of hot reservoir is $99^{\circ} C$ , will be

1

62 K

2

16.5 K

3

33 K

4

66 K

Explanation:

A
Let $T_{2}$ is the temperature of sink and $T_{1}$ is the temperature of source,
Given, $T_{1} = 99^{\circ} C + 273 = 372 K$
$η_{1} = \frac{1}{3}, η_{2} = \frac{1}{6}$
$η_{1} = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{372}$
$\frac{1}{3} = 1 - \frac{T_{2}}{372}$
$\frac{T_{2}}{372} = 1 - \frac{1}{3} = \frac{2}{3}$
$T_{2} = 248 K (sink temperature)$
Now the sink temperature $(T_{2} = 248 K)$ is increased by $x$ $&$ efficiency become $\frac{1}{6}$ ,
$η_{2} = 1 - \frac{T_{2} + x}{T_{1}}$
$\frac{1}{6} = 1 - \frac{248 + x}{372}$
$\frac{248 + x}{372} = 1 - \frac{1}{6} = \frac{5}{6}$
$248 + x = \frac{5}{6} \times 372$
$248 + x = 5 \times 62$
$248 + x = 310$
$x = 62 K$

JEE Main-01.02.2023

Thermodynamics

148520 A Carnot engine whose low temperature reservoir is at $350 K$ has an efficiency of $50 %$ It is desired to increase this to $60 %$ . If the temperature of the low temperature reservoir remains constant, then the temperature of high temperature reservoir must be increased by how many degrees?

1 15

2 175

3 100

4 50

5 120

Explanation:

B Efficiency of Carnot engine-
$η = 1 - \frac{T_{cold}}{T_{hot}}$
When, $η = 50 %$
$\frac{50}{100} = 1 - \frac{T_{cold}}{T_{hot}}$
$0.5 = 1 - \frac{350}{T_{hot}} \Rightarrow T_{hot} = 700 K$
When, $η = 60 %$
$\frac{60}{100} = 1 - \frac{T_{cold}}{T_{hot}^{'}}$
$\frac{60}{100} = 1 - \frac{350}{T_{hot}^{'}}$
$T_{hot}^{'} = 875 K$
Hence the temperature of high reservoir is increased by-
$T_{hot}^{'} - T_{hot} = 875 K - 700 K$
$= 175 K or 175^{\circ} C$

Kerala CEE -2018

Thermodynamics

148522 A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency 0.2 . When $T_{2}$ is reduced by $50 K$ , its efficiency increases to 0.4 . Then, $T_{1}$ and $T_{2}$ are respectively

1

200 K, 150 K

2

250 K, 200 K

3

300 K, 250 K

4

300 K, 200 K

5

300 K, 150 K

Explanation:

B By Carnot's ideal heat engine-
$η = 1 - \frac{T_{2}}{T_{1}}$
Where, $η_{1} = 0.2 & η_{2} = 0.4$
For the first condition-
$η_{1} = 1 - \frac{T_{2}}{T_{1}} \Rightarrow 0.2 = 1 - \frac{T_{2}}{T_{1}}$
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
For the second condition-
$η_{2} = 1 - \frac{T_{2} - 50}{T_{1}}$
$0.4 = \frac{T_{1} - (T_{2} - 50)}{T_{1}}$
$0.4 = \frac{T_{1} - T_{2} + 50}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}} + \frac{50}{T_{1}}$
$0.4 = 0.2 + \frac{50}{T_{1}}$
$0.2 = \frac{50}{T_{1}}$
$T_{1} = 250 K$
Put the value of $T_{1}$ in equation (i), we get-
$0.2 = \frac{T_{1} - T_{2}}{T_{1}}$
$T_{2} = T_{1} - 0.2 T_{1}$
$T_{2} = 250 - 0.2 \times 250 = 250 - 50$
$T_{2} = 200 K$

Kerala CEE- 2014

Thermodynamics

148524 A Carnot engine whose efficiency is $40 %$ , receives heat at $500 K$ . If the efficiency is to be $50 %$ , the source temperature for the same exhaust temperature is

1

900 K

2

600 K

3

700 K

4

800 K

5

550 K

Explanation:

B Carnot efficiency relation-
$η = \frac{T_{1} - T_{2}}{T_{1}}$
Case-1 $\frac{T_{1} - T_{2}}{T_{1}} = 0.4$
$T_{1} - T_{2} = 0.4 T_{1}$
$T_{2} = 0.6 T_{1}$
Case-2 $\frac{T_{1}^{'} - T_{2}}{T_{1}^{'}} = 0.5$
$T_{1}^{'} = \frac{0.6}{0.5} T_{1} = \frac{6}{5} \times 500$
$T_{1}^{'} = 600 K$

Kerala CEE - 2011