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Thermodynamics

148647 When the gas expands with temperature using the relation $V = {KT}^{2 / 3}$ for the temperature change of $40 K$ , the work done is

1

20.1 R

2

30.2 R

3

26.6 R

4

35.6 R

Explanation:

C Given that,
$V = {KT}^{2 / 3}$
$dV = K \frac{2}{3} T^{\frac{- 1}{3}} dT$
Temperature change $(Δ T) = 40 K$
Work done, $(dW) = PdV$
From ideal gas equation,
$PV = nRT$
$P = \frac{nRT}{V} = \frac{nRT}{{KT}^{\frac{2}{3}}}$
$dW = P \cdot Δ V$
$dW = \frac{nRT}{{KT}^{2 / 3}} \cdot K \frac{2}{3} T^{\frac{- 1}{3}} dT$
$= \frac{2 nR}{3} (T^{1 - \frac{1}{3} - \frac{2}{3}}) dT$
$= \frac{2}{3} nR \int_{0}^{40} dT (∵ n = 1)$
$∴ dW = \frac{2}{3} R [T]_{0}^{40}$
$= \frac{2}{3} R (40 - 0)$
$= \frac{80}{3} R$
$= 26.66 R$
$dW = 26.67 R$

J and K-CET-2014

Thermodynamics

148649 Consider a reversible engine of efficiency $\frac{1}{6}$ .
When the temperature of the sink is reduced by $62^{\circ} C$ , its efficiency gets doubled. The temperature of the source and sink respectively are.

1

372 K

and

310 K

2

273 K

and

300 K

3

99^{\circ} C

and

10^{\circ} C

4

200^{\circ} C

and

37^{\circ} C

Explanation:

A Given that,
Initial efficiency of reversible engine $= \frac{1}{6}$
Decreasing temperature $= 62^{\circ} C$
Final efficiency $= 2 \times \frac{1}{6} = \frac{1}{3}$ .
As we know that, $η = \frac{T_{H} - T_{L}}{T_{H}}$
$\frac{1}{6} = 1 - \frac{T_{L}}{T_{H}}$
According to question,
$\frac{1}{3} = 1 - \frac{(T_{L} - 62)}{T_{H}}$
$= \frac{T_{H} - T_{L} + 62}{T_{H}}$
$\frac{1}{3} = \frac{T_{H} - T_{L}}{T_{H}} + \frac{62}{T_{H}}$
From equation (i), we get
$\frac{1}{3} = \frac{1}{6} + \frac{62}{T_{H}}$
$\frac{1}{3} - \frac{1}{6} = \frac{62}{T_{H}}$
$\frac{1}{6} = \frac{62}{T_{H}}$
$T_{H} = 372 K$
Putting the value of $T_{H}$ in equation (i), we get
$\frac{1}{6} = 1 - \frac{T_{L}}{372}$
$\frac{T_{L}}{372} = \frac{5}{6}$
$T_{L} = \frac{5 \times 372}{6}$
$T_{L} = 5 \times 62$
$T_{L} = 310 K$

TS EAMCET (Engg.)-2017

Thermodynamics

148650 A Carnot engine working between $200 K$ and $500 K$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is

1

\frac{4000}{3} J

2

\frac{2000}{3} J

3

\frac{800}{3} J

4

\frac{1600}{3} J

Explanation:

A Given that,
Initial temperature $(T_{L}) = 200 K$
Final temperature, $(T_{H}) = 500 K$
Work done $= 800$ Joule
Efficiency, $η = \frac{T_{H} - T_{L}}{T_{H}} = \frac{Work done}{Heat supply}$
$\frac{500 - 200}{500} = \frac{800}{Heat supply}$
$\frac{300}{500} = \frac{800}{Heat supply}$
Heat supply $= \frac{800 \times 5}{3}$
$= \frac{4000}{3}$

TS EAMCET(Medical)-2015

Thermodynamics

148651 Two Carnot engines $A$ and $B$ are connected in series in such a way that the work outputs are equal when the temperatures of hot and cold reservoirs of $A$ are $800 K$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is

1

400 K

2

450 K

3

500 K

4

550 K

Explanation:

D Given, Initial temperature of $A = 800 K$
Final temperature of $B = 300 K$

Work done for engine $A =$ Work done for engine $B$
$W_{A} = W_{B}$
So, $800 - T = T - 300$
$2 T = 300 + 800$
$2 T = 1100$
$T = \frac{1100}{2}$
$T = 550 K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148652 A Carnot engine has efficiency of $50 %$ . If the temperature of sink is reduced by $40^{\circ} C$ , its efficiency increases by $30 %$ . The temperature of the source will be:

1

166.7 K

2

255.1 K

3

266.7 K

4

367.7 K

Explanation:

C Given that,
Initial efficiency $(η_{1}) = 50 % = 0.5$
Reduced temperature $= 40^{\circ} C$
Increased efficiency $= 30 %$
New efficiency $(η_{2}) = η_{1} \times 1.3 = 0.5 \times 1.3$
$η_{2} = 0.65$ Initial efficiency,
$η_{1} = 1 - \frac{T_{L}}{T_{H}}$
$0.5 = 1 - \frac{T_{L}}{T_{H}}$
$\frac{T_{L}}{T_{H}} = \frac{1}{2}$
$T_{H} = 2 T_{L}$
New efficiency,
$η_{2} = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.65 = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.35 = \frac{(T_{L} - 40)}{T_{H}}$
$0.35 \times 2 T_{L} = T_{L} - 40 (∵ T_{H} = 2 T_{L})$
$0.70 T_{L} = T_{L} - 40$
$T_{L} = \frac{40}{0.30}$
$T_{L} = 133.33 K$
$∴ \frac{T_{H}}{2} = 133.33$
$T_{H} = 266.67 K$

JEE Main-28.07.2022

Thermodynamics

148647 When the gas expands with temperature using the relation $V = {KT}^{2 / 3}$ for the temperature change of $40 K$ , the work done is

1

20.1 R

2

30.2 R

3

26.6 R

4

35.6 R

Explanation:

C Given that,
$V = {KT}^{2 / 3}$
$dV = K \frac{2}{3} T^{\frac{- 1}{3}} dT$
Temperature change $(Δ T) = 40 K$
Work done, $(dW) = PdV$
From ideal gas equation,
$PV = nRT$
$P = \frac{nRT}{V} = \frac{nRT}{{KT}^{\frac{2}{3}}}$
$dW = P \cdot Δ V$
$dW = \frac{nRT}{{KT}^{2 / 3}} \cdot K \frac{2}{3} T^{\frac{- 1}{3}} dT$
$= \frac{2 nR}{3} (T^{1 - \frac{1}{3} - \frac{2}{3}}) dT$
$= \frac{2}{3} nR \int_{0}^{40} dT (∵ n = 1)$
$∴ dW = \frac{2}{3} R [T]_{0}^{40}$
$= \frac{2}{3} R (40 - 0)$
$= \frac{80}{3} R$
$= 26.66 R$
$dW = 26.67 R$

J and K-CET-2014

Thermodynamics

148649 Consider a reversible engine of efficiency $\frac{1}{6}$ .
When the temperature of the sink is reduced by $62^{\circ} C$ , its efficiency gets doubled. The temperature of the source and sink respectively are.

1

372 K

and

310 K

2

273 K

and

300 K

3

99^{\circ} C

and

10^{\circ} C

4

200^{\circ} C

and

37^{\circ} C

Explanation:

A Given that,
Initial efficiency of reversible engine $= \frac{1}{6}$
Decreasing temperature $= 62^{\circ} C$
Final efficiency $= 2 \times \frac{1}{6} = \frac{1}{3}$ .
As we know that, $η = \frac{T_{H} - T_{L}}{T_{H}}$
$\frac{1}{6} = 1 - \frac{T_{L}}{T_{H}}$
According to question,
$\frac{1}{3} = 1 - \frac{(T_{L} - 62)}{T_{H}}$
$= \frac{T_{H} - T_{L} + 62}{T_{H}}$
$\frac{1}{3} = \frac{T_{H} - T_{L}}{T_{H}} + \frac{62}{T_{H}}$
From equation (i), we get
$\frac{1}{3} = \frac{1}{6} + \frac{62}{T_{H}}$
$\frac{1}{3} - \frac{1}{6} = \frac{62}{T_{H}}$
$\frac{1}{6} = \frac{62}{T_{H}}$
$T_{H} = 372 K$
Putting the value of $T_{H}$ in equation (i), we get
$\frac{1}{6} = 1 - \frac{T_{L}}{372}$
$\frac{T_{L}}{372} = \frac{5}{6}$
$T_{L} = \frac{5 \times 372}{6}$
$T_{L} = 5 \times 62$
$T_{L} = 310 K$

TS EAMCET (Engg.)-2017

Thermodynamics

148650 A Carnot engine working between $200 K$ and $500 K$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is

1

\frac{4000}{3} J

2

\frac{2000}{3} J

3

\frac{800}{3} J

4

\frac{1600}{3} J

Explanation:

A Given that,
Initial temperature $(T_{L}) = 200 K$
Final temperature, $(T_{H}) = 500 K$
Work done $= 800$ Joule
Efficiency, $η = \frac{T_{H} - T_{L}}{T_{H}} = \frac{Work done}{Heat supply}$
$\frac{500 - 200}{500} = \frac{800}{Heat supply}$
$\frac{300}{500} = \frac{800}{Heat supply}$
Heat supply $= \frac{800 \times 5}{3}$
$= \frac{4000}{3}$

TS EAMCET(Medical)-2015

Thermodynamics

148651 Two Carnot engines $A$ and $B$ are connected in series in such a way that the work outputs are equal when the temperatures of hot and cold reservoirs of $A$ are $800 K$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is

1

400 K

2

450 K

3

500 K

4

550 K

Explanation:

D Given, Initial temperature of $A = 800 K$
Final temperature of $B = 300 K$

Work done for engine $A =$ Work done for engine $B$
$W_{A} = W_{B}$
So, $800 - T = T - 300$
$2 T = 300 + 800$
$2 T = 1100$
$T = \frac{1100}{2}$
$T = 550 K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148652 A Carnot engine has efficiency of $50 %$ . If the temperature of sink is reduced by $40^{\circ} C$ , its efficiency increases by $30 %$ . The temperature of the source will be:

1

166.7 K

2

255.1 K

3

266.7 K

4

367.7 K

Explanation:

C Given that,
Initial efficiency $(η_{1}) = 50 % = 0.5$
Reduced temperature $= 40^{\circ} C$
Increased efficiency $= 30 %$
New efficiency $(η_{2}) = η_{1} \times 1.3 = 0.5 \times 1.3$
$η_{2} = 0.65$ Initial efficiency,
$η_{1} = 1 - \frac{T_{L}}{T_{H}}$
$0.5 = 1 - \frac{T_{L}}{T_{H}}$
$\frac{T_{L}}{T_{H}} = \frac{1}{2}$
$T_{H} = 2 T_{L}$
New efficiency,
$η_{2} = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.65 = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.35 = \frac{(T_{L} - 40)}{T_{H}}$
$0.35 \times 2 T_{L} = T_{L} - 40 (∵ T_{H} = 2 T_{L})$
$0.70 T_{L} = T_{L} - 40$
$T_{L} = \frac{40}{0.30}$
$T_{L} = 133.33 K$
$∴ \frac{T_{H}}{2} = 133.33$
$T_{H} = 266.67 K$

JEE Main-28.07.2022

Thermodynamics

148647 When the gas expands with temperature using the relation $V = {KT}^{2 / 3}$ for the temperature change of $40 K$ , the work done is

1

20.1 R

2

30.2 R

3

26.6 R

4

35.6 R

Explanation:

C Given that,
$V = {KT}^{2 / 3}$
$dV = K \frac{2}{3} T^{\frac{- 1}{3}} dT$
Temperature change $(Δ T) = 40 K$
Work done, $(dW) = PdV$
From ideal gas equation,
$PV = nRT$
$P = \frac{nRT}{V} = \frac{nRT}{{KT}^{\frac{2}{3}}}$
$dW = P \cdot Δ V$
$dW = \frac{nRT}{{KT}^{2 / 3}} \cdot K \frac{2}{3} T^{\frac{- 1}{3}} dT$
$= \frac{2 nR}{3} (T^{1 - \frac{1}{3} - \frac{2}{3}}) dT$
$= \frac{2}{3} nR \int_{0}^{40} dT (∵ n = 1)$
$∴ dW = \frac{2}{3} R [T]_{0}^{40}$
$= \frac{2}{3} R (40 - 0)$
$= \frac{80}{3} R$
$= 26.66 R$
$dW = 26.67 R$

J and K-CET-2014

Thermodynamics

148649 Consider a reversible engine of efficiency $\frac{1}{6}$ .
When the temperature of the sink is reduced by $62^{\circ} C$ , its efficiency gets doubled. The temperature of the source and sink respectively are.

1

372 K

and

310 K

2

273 K

and

300 K

3

99^{\circ} C

and

10^{\circ} C

4

200^{\circ} C

and

37^{\circ} C

Explanation:

A Given that,
Initial efficiency of reversible engine $= \frac{1}{6}$
Decreasing temperature $= 62^{\circ} C$
Final efficiency $= 2 \times \frac{1}{6} = \frac{1}{3}$ .
As we know that, $η = \frac{T_{H} - T_{L}}{T_{H}}$
$\frac{1}{6} = 1 - \frac{T_{L}}{T_{H}}$
According to question,
$\frac{1}{3} = 1 - \frac{(T_{L} - 62)}{T_{H}}$
$= \frac{T_{H} - T_{L} + 62}{T_{H}}$
$\frac{1}{3} = \frac{T_{H} - T_{L}}{T_{H}} + \frac{62}{T_{H}}$
From equation (i), we get
$\frac{1}{3} = \frac{1}{6} + \frac{62}{T_{H}}$
$\frac{1}{3} - \frac{1}{6} = \frac{62}{T_{H}}$
$\frac{1}{6} = \frac{62}{T_{H}}$
$T_{H} = 372 K$
Putting the value of $T_{H}$ in equation (i), we get
$\frac{1}{6} = 1 - \frac{T_{L}}{372}$
$\frac{T_{L}}{372} = \frac{5}{6}$
$T_{L} = \frac{5 \times 372}{6}$
$T_{L} = 5 \times 62$
$T_{L} = 310 K$

TS EAMCET (Engg.)-2017

Thermodynamics

148650 A Carnot engine working between $200 K$ and $500 K$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is

1

\frac{4000}{3} J

2

\frac{2000}{3} J

3

\frac{800}{3} J

4

\frac{1600}{3} J

Explanation:

A Given that,
Initial temperature $(T_{L}) = 200 K$
Final temperature, $(T_{H}) = 500 K$
Work done $= 800$ Joule
Efficiency, $η = \frac{T_{H} - T_{L}}{T_{H}} = \frac{Work done}{Heat supply}$
$\frac{500 - 200}{500} = \frac{800}{Heat supply}$
$\frac{300}{500} = \frac{800}{Heat supply}$
Heat supply $= \frac{800 \times 5}{3}$
$= \frac{4000}{3}$

TS EAMCET(Medical)-2015

Thermodynamics

148651 Two Carnot engines $A$ and $B$ are connected in series in such a way that the work outputs are equal when the temperatures of hot and cold reservoirs of $A$ are $800 K$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is

1

400 K

2

450 K

3

500 K

4

550 K

Explanation:

D Given, Initial temperature of $A = 800 K$
Final temperature of $B = 300 K$

Work done for engine $A =$ Work done for engine $B$
$W_{A} = W_{B}$
So, $800 - T = T - 300$
$2 T = 300 + 800$
$2 T = 1100$
$T = \frac{1100}{2}$
$T = 550 K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148652 A Carnot engine has efficiency of $50 %$ . If the temperature of sink is reduced by $40^{\circ} C$ , its efficiency increases by $30 %$ . The temperature of the source will be:

1

166.7 K

2

255.1 K

3

266.7 K

4

367.7 K

Explanation:

C Given that,
Initial efficiency $(η_{1}) = 50 % = 0.5$
Reduced temperature $= 40^{\circ} C$
Increased efficiency $= 30 %$
New efficiency $(η_{2}) = η_{1} \times 1.3 = 0.5 \times 1.3$
$η_{2} = 0.65$ Initial efficiency,
$η_{1} = 1 - \frac{T_{L}}{T_{H}}$
$0.5 = 1 - \frac{T_{L}}{T_{H}}$
$\frac{T_{L}}{T_{H}} = \frac{1}{2}$
$T_{H} = 2 T_{L}$
New efficiency,
$η_{2} = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.65 = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.35 = \frac{(T_{L} - 40)}{T_{H}}$
$0.35 \times 2 T_{L} = T_{L} - 40 (∵ T_{H} = 2 T_{L})$
$0.70 T_{L} = T_{L} - 40$
$T_{L} = \frac{40}{0.30}$
$T_{L} = 133.33 K$
$∴ \frac{T_{H}}{2} = 133.33$
$T_{H} = 266.67 K$

JEE Main-28.07.2022

Thermodynamics

148647 When the gas expands with temperature using the relation $V = {KT}^{2 / 3}$ for the temperature change of $40 K$ , the work done is

1

20.1 R

2

30.2 R

3

26.6 R

4

35.6 R

Explanation:

C Given that,
$V = {KT}^{2 / 3}$
$dV = K \frac{2}{3} T^{\frac{- 1}{3}} dT$
Temperature change $(Δ T) = 40 K$
Work done, $(dW) = PdV$
From ideal gas equation,
$PV = nRT$
$P = \frac{nRT}{V} = \frac{nRT}{{KT}^{\frac{2}{3}}}$
$dW = P \cdot Δ V$
$dW = \frac{nRT}{{KT}^{2 / 3}} \cdot K \frac{2}{3} T^{\frac{- 1}{3}} dT$
$= \frac{2 nR}{3} (T^{1 - \frac{1}{3} - \frac{2}{3}}) dT$
$= \frac{2}{3} nR \int_{0}^{40} dT (∵ n = 1)$
$∴ dW = \frac{2}{3} R [T]_{0}^{40}$
$= \frac{2}{3} R (40 - 0)$
$= \frac{80}{3} R$
$= 26.66 R$
$dW = 26.67 R$

J and K-CET-2014

Thermodynamics

148649 Consider a reversible engine of efficiency $\frac{1}{6}$ .
When the temperature of the sink is reduced by $62^{\circ} C$ , its efficiency gets doubled. The temperature of the source and sink respectively are.

1

372 K

and

310 K

2

273 K

and

300 K

3

99^{\circ} C

and

10^{\circ} C

4

200^{\circ} C

and

37^{\circ} C

Explanation:

A Given that,
Initial efficiency of reversible engine $= \frac{1}{6}$
Decreasing temperature $= 62^{\circ} C$
Final efficiency $= 2 \times \frac{1}{6} = \frac{1}{3}$ .
As we know that, $η = \frac{T_{H} - T_{L}}{T_{H}}$
$\frac{1}{6} = 1 - \frac{T_{L}}{T_{H}}$
According to question,
$\frac{1}{3} = 1 - \frac{(T_{L} - 62)}{T_{H}}$
$= \frac{T_{H} - T_{L} + 62}{T_{H}}$
$\frac{1}{3} = \frac{T_{H} - T_{L}}{T_{H}} + \frac{62}{T_{H}}$
From equation (i), we get
$\frac{1}{3} = \frac{1}{6} + \frac{62}{T_{H}}$
$\frac{1}{3} - \frac{1}{6} = \frac{62}{T_{H}}$
$\frac{1}{6} = \frac{62}{T_{H}}$
$T_{H} = 372 K$
Putting the value of $T_{H}$ in equation (i), we get
$\frac{1}{6} = 1 - \frac{T_{L}}{372}$
$\frac{T_{L}}{372} = \frac{5}{6}$
$T_{L} = \frac{5 \times 372}{6}$
$T_{L} = 5 \times 62$
$T_{L} = 310 K$

TS EAMCET (Engg.)-2017

Thermodynamics

148650 A Carnot engine working between $200 K$ and $500 K$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is

1

\frac{4000}{3} J

2

\frac{2000}{3} J

3

\frac{800}{3} J

4

\frac{1600}{3} J

Explanation:

A Given that,
Initial temperature $(T_{L}) = 200 K$
Final temperature, $(T_{H}) = 500 K$
Work done $= 800$ Joule
Efficiency, $η = \frac{T_{H} - T_{L}}{T_{H}} = \frac{Work done}{Heat supply}$
$\frac{500 - 200}{500} = \frac{800}{Heat supply}$
$\frac{300}{500} = \frac{800}{Heat supply}$
Heat supply $= \frac{800 \times 5}{3}$
$= \frac{4000}{3}$

TS EAMCET(Medical)-2015

Thermodynamics

148651 Two Carnot engines $A$ and $B$ are connected in series in such a way that the work outputs are equal when the temperatures of hot and cold reservoirs of $A$ are $800 K$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is

1

400 K

2

450 K

3

500 K

4

550 K

Explanation:

D Given, Initial temperature of $A = 800 K$
Final temperature of $B = 300 K$

Work done for engine $A =$ Work done for engine $B$
$W_{A} = W_{B}$
So, $800 - T = T - 300$
$2 T = 300 + 800$
$2 T = 1100$
$T = \frac{1100}{2}$
$T = 550 K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148652 A Carnot engine has efficiency of $50 %$ . If the temperature of sink is reduced by $40^{\circ} C$ , its efficiency increases by $30 %$ . The temperature of the source will be:

1

166.7 K

2

255.1 K

3

266.7 K

4

367.7 K

Explanation:

C Given that,
Initial efficiency $(η_{1}) = 50 % = 0.5$
Reduced temperature $= 40^{\circ} C$
Increased efficiency $= 30 %$
New efficiency $(η_{2}) = η_{1} \times 1.3 = 0.5 \times 1.3$
$η_{2} = 0.65$ Initial efficiency,
$η_{1} = 1 - \frac{T_{L}}{T_{H}}$
$0.5 = 1 - \frac{T_{L}}{T_{H}}$
$\frac{T_{L}}{T_{H}} = \frac{1}{2}$
$T_{H} = 2 T_{L}$
New efficiency,
$η_{2} = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.65 = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.35 = \frac{(T_{L} - 40)}{T_{H}}$
$0.35 \times 2 T_{L} = T_{L} - 40 (∵ T_{H} = 2 T_{L})$
$0.70 T_{L} = T_{L} - 40$
$T_{L} = \frac{40}{0.30}$
$T_{L} = 133.33 K$
$∴ \frac{T_{H}}{2} = 133.33$
$T_{H} = 266.67 K$

JEE Main-28.07.2022

Thermodynamics

148647 When the gas expands with temperature using the relation $V = {KT}^{2 / 3}$ for the temperature change of $40 K$ , the work done is

1

20.1 R

2

30.2 R

3

26.6 R

4

35.6 R

Explanation:

C Given that,
$V = {KT}^{2 / 3}$
$dV = K \frac{2}{3} T^{\frac{- 1}{3}} dT$
Temperature change $(Δ T) = 40 K$
Work done, $(dW) = PdV$
From ideal gas equation,
$PV = nRT$
$P = \frac{nRT}{V} = \frac{nRT}{{KT}^{\frac{2}{3}}}$
$dW = P \cdot Δ V$
$dW = \frac{nRT}{{KT}^{2 / 3}} \cdot K \frac{2}{3} T^{\frac{- 1}{3}} dT$
$= \frac{2 nR}{3} (T^{1 - \frac{1}{3} - \frac{2}{3}}) dT$
$= \frac{2}{3} nR \int_{0}^{40} dT (∵ n = 1)$
$∴ dW = \frac{2}{3} R [T]_{0}^{40}$
$= \frac{2}{3} R (40 - 0)$
$= \frac{80}{3} R$
$= 26.66 R$
$dW = 26.67 R$

J and K-CET-2014

Thermodynamics

148649 Consider a reversible engine of efficiency $\frac{1}{6}$ .
When the temperature of the sink is reduced by $62^{\circ} C$ , its efficiency gets doubled. The temperature of the source and sink respectively are.

1

372 K

and

310 K

2

273 K

and

300 K

3

99^{\circ} C

and

10^{\circ} C

4

200^{\circ} C

and

37^{\circ} C

Explanation:

A Given that,
Initial efficiency of reversible engine $= \frac{1}{6}$
Decreasing temperature $= 62^{\circ} C$
Final efficiency $= 2 \times \frac{1}{6} = \frac{1}{3}$ .
As we know that, $η = \frac{T_{H} - T_{L}}{T_{H}}$
$\frac{1}{6} = 1 - \frac{T_{L}}{T_{H}}$
According to question,
$\frac{1}{3} = 1 - \frac{(T_{L} - 62)}{T_{H}}$
$= \frac{T_{H} - T_{L} + 62}{T_{H}}$
$\frac{1}{3} = \frac{T_{H} - T_{L}}{T_{H}} + \frac{62}{T_{H}}$
From equation (i), we get
$\frac{1}{3} = \frac{1}{6} + \frac{62}{T_{H}}$
$\frac{1}{3} - \frac{1}{6} = \frac{62}{T_{H}}$
$\frac{1}{6} = \frac{62}{T_{H}}$
$T_{H} = 372 K$
Putting the value of $T_{H}$ in equation (i), we get
$\frac{1}{6} = 1 - \frac{T_{L}}{372}$
$\frac{T_{L}}{372} = \frac{5}{6}$
$T_{L} = \frac{5 \times 372}{6}$
$T_{L} = 5 \times 62$
$T_{L} = 310 K$

TS EAMCET (Engg.)-2017

Thermodynamics

148650 A Carnot engine working between $200 K$ and $500 K$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is

1

\frac{4000}{3} J

2

\frac{2000}{3} J

3

\frac{800}{3} J

4

\frac{1600}{3} J

Explanation:

A Given that,
Initial temperature $(T_{L}) = 200 K$
Final temperature, $(T_{H}) = 500 K$
Work done $= 800$ Joule
Efficiency, $η = \frac{T_{H} - T_{L}}{T_{H}} = \frac{Work done}{Heat supply}$
$\frac{500 - 200}{500} = \frac{800}{Heat supply}$
$\frac{300}{500} = \frac{800}{Heat supply}$
Heat supply $= \frac{800 \times 5}{3}$
$= \frac{4000}{3}$

TS EAMCET(Medical)-2015

Thermodynamics

148651 Two Carnot engines $A$ and $B$ are connected in series in such a way that the work outputs are equal when the temperatures of hot and cold reservoirs of $A$ are $800 K$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is

1

400 K

2

450 K

3

500 K

4

550 K

Explanation:

D Given, Initial temperature of $A = 800 K$
Final temperature of $B = 300 K$

Work done for engine $A =$ Work done for engine $B$
$W_{A} = W_{B}$
So, $800 - T = T - 300$
$2 T = 300 + 800$
$2 T = 1100$
$T = \frac{1100}{2}$
$T = 550 K$

TS EAMCET (Medical)-02.05.2018

Thermodynamics

148652 A Carnot engine has efficiency of $50 %$ . If the temperature of sink is reduced by $40^{\circ} C$ , its efficiency increases by $30 %$ . The temperature of the source will be:

1

166.7 K

2

255.1 K

3

266.7 K

4

367.7 K

Explanation:

C Given that,
Initial efficiency $(η_{1}) = 50 % = 0.5$
Reduced temperature $= 40^{\circ} C$
Increased efficiency $= 30 %$
New efficiency $(η_{2}) = η_{1} \times 1.3 = 0.5 \times 1.3$
$η_{2} = 0.65$ Initial efficiency,
$η_{1} = 1 - \frac{T_{L}}{T_{H}}$
$0.5 = 1 - \frac{T_{L}}{T_{H}}$
$\frac{T_{L}}{T_{H}} = \frac{1}{2}$
$T_{H} = 2 T_{L}$
New efficiency,
$η_{2} = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.65 = 1 - \frac{(T_{L} - 40)}{T_{H}}$
$0.35 = \frac{(T_{L} - 40)}{T_{H}}$
$0.35 \times 2 T_{L} = T_{L} - 40 (∵ T_{H} = 2 T_{L})$
$0.70 T_{L} = T_{L} - 40$
$T_{L} = \frac{40}{0.30}$
$T_{L} = 133.33 K$
$∴ \frac{T_{H}}{2} = 133.33$
$T_{H} = 266.67 K$

JEE Main-28.07.2022