148647
When the gas expands with temperature using the relation $\mathrm{V}=\mathrm{KT}^{2 / 3}$ for the temperature change of $40 \mathrm{~K}$, the work done is
1 $20.1 \mathrm{R}$
2 $30.2 \mathrm{R}$
3 $26.6 \mathrm{R}$
4 $35.6 \mathrm{R}$
Explanation:
C Given that, $\mathrm{V}=\mathrm{KT}^{2 / 3}$ $\mathrm{dV}=\mathrm{K} \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ Temperature change $(\Delta \mathrm{T})=40 \mathrm{~K}$ Work done, $(\mathrm{dW})=\mathrm{PdV}$ From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\frac{\mathrm{nRT}}{\mathrm{KT}^{\frac{2}{3}}}$ $\mathrm{dW}=\mathrm{P} \cdot \Delta \mathrm{V}$ $\mathrm{dW}=\frac{\mathrm{nRT}}{\mathrm{KT}^{2 / 3}} \cdot K \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ $=\frac{2 \mathrm{nR}}{3}\left(\mathrm{~T}^{1-\frac{1}{3}-\frac{2}{3}}\right) \mathrm{dT}$ $=\frac{2}{3} \mathrm{nR} \int_{0}^{40} \mathrm{dT} \quad(\because \mathrm{n}=1)$ $\therefore \mathrm{dW}=\frac{2}{3} \mathrm{R}[\mathrm{T}]_{0}^{40}$ $=\frac{2}{3} \mathrm{R}(40-0)$ $=\frac{80}{3} \mathrm{R}$ $=26.66 \mathrm{R}$ $\mathrm{dW}=26.67 \mathrm{R}$
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 $62^{\circ} \mathrm{C}$, its efficiency gets doubled. The temperature of the source and sink respectively are.
1 $372 \mathrm{~K}$ and $310 \mathrm{~K}$
2 $273 \mathrm{~K}$ and $300 \mathrm{~K}$
3 $99^{\circ} \mathrm{C}$ and $10^{\circ} \mathrm{C}$
4 $200^{\circ} \mathrm{C}$ and $37^{\circ} \mathrm{C}$
Explanation:
A Given that, Initial efficiency of reversible engine $=\frac{1}{6}$ Decreasing temperature $=62^{\circ} \mathrm{C}$ Final efficiency $=2 \times \frac{1}{6}=\frac{1}{3}$. As we know that, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ According to question, $\frac{1}{3}=1-\frac{\left(\mathrm{T}_{\mathrm{L}}-62\right)}{\mathrm{T}_{\mathrm{H}}}$ $=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}+62}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{1}{3}=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}+\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ From equation (i), we get $\frac{1}{3}=\frac{1}{6}+\frac{62}{T_{H}}$ $\frac{1}{3}-\frac{1}{6}=\frac{62}{T_{H}}$ $\frac{1}{6}=\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}}=372 \mathrm{~K}$ Putting the value of $T_{H}$ in equation (i), we get $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{372}$ $\frac{\mathrm{T}_{\mathrm{L}}}{372}=\frac{5}{6}$ $\mathrm{T}_{\mathrm{L}}=\frac{5 \times 372}{6}$ $\mathrm{T}_{\mathrm{L}}=5 \times 62$ $\mathrm{T}_{\mathrm{L}}=310 \mathrm{~K}$
TS EAMCET (Engg.)-2017
Thermodynamics
148650
A Carnot engine working between $200 \mathrm{~K}$ and $500 \mathrm{~K}$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is
1 $\frac{4000}{3} \mathrm{~J}$
2 $\frac{2000}{3} \mathrm{~J}$
3 $\frac{800}{3} \mathrm{~J}$
4 $\frac{1600}{3} \mathrm{~J}$
Explanation:
A Given that, Initial temperature $\left(\mathrm{T}_{\mathrm{L}}\right)=200 \mathrm{~K}$ Final temperature, $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Work done $=800$ Joule Efficiency, $\eta=\frac{T_{H}-T_{L}}{T_{H}}=\frac{\text { Work done }}{\text { Heat supply }}$ $\frac{500-200}{500}=\frac{800}{\text { Heat supply }}$ $\frac{300}{500}=\frac{800}{\text { Heat supply }}$ 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 $800 \mathrm{~K}$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is
1 $400 \mathrm{~K}$
2 $450 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $550 \mathrm{~K}$
Explanation:
D Given, Initial temperature of $\mathrm{A}=800 \mathrm{~K}$ Final temperature of $B=300 \mathrm{~K}$ Work done for engine $A=$ Work done for engine $B$ $\mathrm{W}_{\mathrm{A}}=\mathrm{W}_{\mathrm{B}}$ So, $\quad 800-\mathrm{T}=\mathrm{T}-300$ $2 \mathrm{~T}=300+800$ $2 \mathrm{~T}=1100$ $\mathrm{T}=\frac{1100}{2}$ $\mathrm{T}=550 \mathrm{~K}$
TS EAMCET (Medical)-02.05.2018
Thermodynamics
148652
A Carnot engine has efficiency of $50 \%$. If the temperature of sink is reduced by $40^{\circ} \mathrm{C}$, its efficiency increases by $30 \%$. The temperature of the source will be:
148647
When the gas expands with temperature using the relation $\mathrm{V}=\mathrm{KT}^{2 / 3}$ for the temperature change of $40 \mathrm{~K}$, the work done is
1 $20.1 \mathrm{R}$
2 $30.2 \mathrm{R}$
3 $26.6 \mathrm{R}$
4 $35.6 \mathrm{R}$
Explanation:
C Given that, $\mathrm{V}=\mathrm{KT}^{2 / 3}$ $\mathrm{dV}=\mathrm{K} \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ Temperature change $(\Delta \mathrm{T})=40 \mathrm{~K}$ Work done, $(\mathrm{dW})=\mathrm{PdV}$ From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\frac{\mathrm{nRT}}{\mathrm{KT}^{\frac{2}{3}}}$ $\mathrm{dW}=\mathrm{P} \cdot \Delta \mathrm{V}$ $\mathrm{dW}=\frac{\mathrm{nRT}}{\mathrm{KT}^{2 / 3}} \cdot K \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ $=\frac{2 \mathrm{nR}}{3}\left(\mathrm{~T}^{1-\frac{1}{3}-\frac{2}{3}}\right) \mathrm{dT}$ $=\frac{2}{3} \mathrm{nR} \int_{0}^{40} \mathrm{dT} \quad(\because \mathrm{n}=1)$ $\therefore \mathrm{dW}=\frac{2}{3} \mathrm{R}[\mathrm{T}]_{0}^{40}$ $=\frac{2}{3} \mathrm{R}(40-0)$ $=\frac{80}{3} \mathrm{R}$ $=26.66 \mathrm{R}$ $\mathrm{dW}=26.67 \mathrm{R}$
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 $62^{\circ} \mathrm{C}$, its efficiency gets doubled. The temperature of the source and sink respectively are.
1 $372 \mathrm{~K}$ and $310 \mathrm{~K}$
2 $273 \mathrm{~K}$ and $300 \mathrm{~K}$
3 $99^{\circ} \mathrm{C}$ and $10^{\circ} \mathrm{C}$
4 $200^{\circ} \mathrm{C}$ and $37^{\circ} \mathrm{C}$
Explanation:
A Given that, Initial efficiency of reversible engine $=\frac{1}{6}$ Decreasing temperature $=62^{\circ} \mathrm{C}$ Final efficiency $=2 \times \frac{1}{6}=\frac{1}{3}$. As we know that, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ According to question, $\frac{1}{3}=1-\frac{\left(\mathrm{T}_{\mathrm{L}}-62\right)}{\mathrm{T}_{\mathrm{H}}}$ $=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}+62}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{1}{3}=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}+\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ From equation (i), we get $\frac{1}{3}=\frac{1}{6}+\frac{62}{T_{H}}$ $\frac{1}{3}-\frac{1}{6}=\frac{62}{T_{H}}$ $\frac{1}{6}=\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}}=372 \mathrm{~K}$ Putting the value of $T_{H}$ in equation (i), we get $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{372}$ $\frac{\mathrm{T}_{\mathrm{L}}}{372}=\frac{5}{6}$ $\mathrm{T}_{\mathrm{L}}=\frac{5 \times 372}{6}$ $\mathrm{T}_{\mathrm{L}}=5 \times 62$ $\mathrm{T}_{\mathrm{L}}=310 \mathrm{~K}$
TS EAMCET (Engg.)-2017
Thermodynamics
148650
A Carnot engine working between $200 \mathrm{~K}$ and $500 \mathrm{~K}$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is
1 $\frac{4000}{3} \mathrm{~J}$
2 $\frac{2000}{3} \mathrm{~J}$
3 $\frac{800}{3} \mathrm{~J}$
4 $\frac{1600}{3} \mathrm{~J}$
Explanation:
A Given that, Initial temperature $\left(\mathrm{T}_{\mathrm{L}}\right)=200 \mathrm{~K}$ Final temperature, $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Work done $=800$ Joule Efficiency, $\eta=\frac{T_{H}-T_{L}}{T_{H}}=\frac{\text { Work done }}{\text { Heat supply }}$ $\frac{500-200}{500}=\frac{800}{\text { Heat supply }}$ $\frac{300}{500}=\frac{800}{\text { Heat supply }}$ 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 $800 \mathrm{~K}$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is
1 $400 \mathrm{~K}$
2 $450 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $550 \mathrm{~K}$
Explanation:
D Given, Initial temperature of $\mathrm{A}=800 \mathrm{~K}$ Final temperature of $B=300 \mathrm{~K}$ Work done for engine $A=$ Work done for engine $B$ $\mathrm{W}_{\mathrm{A}}=\mathrm{W}_{\mathrm{B}}$ So, $\quad 800-\mathrm{T}=\mathrm{T}-300$ $2 \mathrm{~T}=300+800$ $2 \mathrm{~T}=1100$ $\mathrm{T}=\frac{1100}{2}$ $\mathrm{T}=550 \mathrm{~K}$
TS EAMCET (Medical)-02.05.2018
Thermodynamics
148652
A Carnot engine has efficiency of $50 \%$. If the temperature of sink is reduced by $40^{\circ} \mathrm{C}$, its efficiency increases by $30 \%$. The temperature of the source will be:
148647
When the gas expands with temperature using the relation $\mathrm{V}=\mathrm{KT}^{2 / 3}$ for the temperature change of $40 \mathrm{~K}$, the work done is
1 $20.1 \mathrm{R}$
2 $30.2 \mathrm{R}$
3 $26.6 \mathrm{R}$
4 $35.6 \mathrm{R}$
Explanation:
C Given that, $\mathrm{V}=\mathrm{KT}^{2 / 3}$ $\mathrm{dV}=\mathrm{K} \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ Temperature change $(\Delta \mathrm{T})=40 \mathrm{~K}$ Work done, $(\mathrm{dW})=\mathrm{PdV}$ From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\frac{\mathrm{nRT}}{\mathrm{KT}^{\frac{2}{3}}}$ $\mathrm{dW}=\mathrm{P} \cdot \Delta \mathrm{V}$ $\mathrm{dW}=\frac{\mathrm{nRT}}{\mathrm{KT}^{2 / 3}} \cdot K \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ $=\frac{2 \mathrm{nR}}{3}\left(\mathrm{~T}^{1-\frac{1}{3}-\frac{2}{3}}\right) \mathrm{dT}$ $=\frac{2}{3} \mathrm{nR} \int_{0}^{40} \mathrm{dT} \quad(\because \mathrm{n}=1)$ $\therefore \mathrm{dW}=\frac{2}{3} \mathrm{R}[\mathrm{T}]_{0}^{40}$ $=\frac{2}{3} \mathrm{R}(40-0)$ $=\frac{80}{3} \mathrm{R}$ $=26.66 \mathrm{R}$ $\mathrm{dW}=26.67 \mathrm{R}$
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 $62^{\circ} \mathrm{C}$, its efficiency gets doubled. The temperature of the source and sink respectively are.
1 $372 \mathrm{~K}$ and $310 \mathrm{~K}$
2 $273 \mathrm{~K}$ and $300 \mathrm{~K}$
3 $99^{\circ} \mathrm{C}$ and $10^{\circ} \mathrm{C}$
4 $200^{\circ} \mathrm{C}$ and $37^{\circ} \mathrm{C}$
Explanation:
A Given that, Initial efficiency of reversible engine $=\frac{1}{6}$ Decreasing temperature $=62^{\circ} \mathrm{C}$ Final efficiency $=2 \times \frac{1}{6}=\frac{1}{3}$. As we know that, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ According to question, $\frac{1}{3}=1-\frac{\left(\mathrm{T}_{\mathrm{L}}-62\right)}{\mathrm{T}_{\mathrm{H}}}$ $=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}+62}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{1}{3}=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}+\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ From equation (i), we get $\frac{1}{3}=\frac{1}{6}+\frac{62}{T_{H}}$ $\frac{1}{3}-\frac{1}{6}=\frac{62}{T_{H}}$ $\frac{1}{6}=\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}}=372 \mathrm{~K}$ Putting the value of $T_{H}$ in equation (i), we get $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{372}$ $\frac{\mathrm{T}_{\mathrm{L}}}{372}=\frac{5}{6}$ $\mathrm{T}_{\mathrm{L}}=\frac{5 \times 372}{6}$ $\mathrm{T}_{\mathrm{L}}=5 \times 62$ $\mathrm{T}_{\mathrm{L}}=310 \mathrm{~K}$
TS EAMCET (Engg.)-2017
Thermodynamics
148650
A Carnot engine working between $200 \mathrm{~K}$ and $500 \mathrm{~K}$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is
1 $\frac{4000}{3} \mathrm{~J}$
2 $\frac{2000}{3} \mathrm{~J}$
3 $\frac{800}{3} \mathrm{~J}$
4 $\frac{1600}{3} \mathrm{~J}$
Explanation:
A Given that, Initial temperature $\left(\mathrm{T}_{\mathrm{L}}\right)=200 \mathrm{~K}$ Final temperature, $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Work done $=800$ Joule Efficiency, $\eta=\frac{T_{H}-T_{L}}{T_{H}}=\frac{\text { Work done }}{\text { Heat supply }}$ $\frac{500-200}{500}=\frac{800}{\text { Heat supply }}$ $\frac{300}{500}=\frac{800}{\text { Heat supply }}$ 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 $800 \mathrm{~K}$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is
1 $400 \mathrm{~K}$
2 $450 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $550 \mathrm{~K}$
Explanation:
D Given, Initial temperature of $\mathrm{A}=800 \mathrm{~K}$ Final temperature of $B=300 \mathrm{~K}$ Work done for engine $A=$ Work done for engine $B$ $\mathrm{W}_{\mathrm{A}}=\mathrm{W}_{\mathrm{B}}$ So, $\quad 800-\mathrm{T}=\mathrm{T}-300$ $2 \mathrm{~T}=300+800$ $2 \mathrm{~T}=1100$ $\mathrm{T}=\frac{1100}{2}$ $\mathrm{T}=550 \mathrm{~K}$
TS EAMCET (Medical)-02.05.2018
Thermodynamics
148652
A Carnot engine has efficiency of $50 \%$. If the temperature of sink is reduced by $40^{\circ} \mathrm{C}$, its efficiency increases by $30 \%$. The temperature of the source will be:
148647
When the gas expands with temperature using the relation $\mathrm{V}=\mathrm{KT}^{2 / 3}$ for the temperature change of $40 \mathrm{~K}$, the work done is
1 $20.1 \mathrm{R}$
2 $30.2 \mathrm{R}$
3 $26.6 \mathrm{R}$
4 $35.6 \mathrm{R}$
Explanation:
C Given that, $\mathrm{V}=\mathrm{KT}^{2 / 3}$ $\mathrm{dV}=\mathrm{K} \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ Temperature change $(\Delta \mathrm{T})=40 \mathrm{~K}$ Work done, $(\mathrm{dW})=\mathrm{PdV}$ From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\frac{\mathrm{nRT}}{\mathrm{KT}^{\frac{2}{3}}}$ $\mathrm{dW}=\mathrm{P} \cdot \Delta \mathrm{V}$ $\mathrm{dW}=\frac{\mathrm{nRT}}{\mathrm{KT}^{2 / 3}} \cdot K \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ $=\frac{2 \mathrm{nR}}{3}\left(\mathrm{~T}^{1-\frac{1}{3}-\frac{2}{3}}\right) \mathrm{dT}$ $=\frac{2}{3} \mathrm{nR} \int_{0}^{40} \mathrm{dT} \quad(\because \mathrm{n}=1)$ $\therefore \mathrm{dW}=\frac{2}{3} \mathrm{R}[\mathrm{T}]_{0}^{40}$ $=\frac{2}{3} \mathrm{R}(40-0)$ $=\frac{80}{3} \mathrm{R}$ $=26.66 \mathrm{R}$ $\mathrm{dW}=26.67 \mathrm{R}$
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 $62^{\circ} \mathrm{C}$, its efficiency gets doubled. The temperature of the source and sink respectively are.
1 $372 \mathrm{~K}$ and $310 \mathrm{~K}$
2 $273 \mathrm{~K}$ and $300 \mathrm{~K}$
3 $99^{\circ} \mathrm{C}$ and $10^{\circ} \mathrm{C}$
4 $200^{\circ} \mathrm{C}$ and $37^{\circ} \mathrm{C}$
Explanation:
A Given that, Initial efficiency of reversible engine $=\frac{1}{6}$ Decreasing temperature $=62^{\circ} \mathrm{C}$ Final efficiency $=2 \times \frac{1}{6}=\frac{1}{3}$. As we know that, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ According to question, $\frac{1}{3}=1-\frac{\left(\mathrm{T}_{\mathrm{L}}-62\right)}{\mathrm{T}_{\mathrm{H}}}$ $=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}+62}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{1}{3}=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}+\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ From equation (i), we get $\frac{1}{3}=\frac{1}{6}+\frac{62}{T_{H}}$ $\frac{1}{3}-\frac{1}{6}=\frac{62}{T_{H}}$ $\frac{1}{6}=\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}}=372 \mathrm{~K}$ Putting the value of $T_{H}$ in equation (i), we get $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{372}$ $\frac{\mathrm{T}_{\mathrm{L}}}{372}=\frac{5}{6}$ $\mathrm{T}_{\mathrm{L}}=\frac{5 \times 372}{6}$ $\mathrm{T}_{\mathrm{L}}=5 \times 62$ $\mathrm{T}_{\mathrm{L}}=310 \mathrm{~K}$
TS EAMCET (Engg.)-2017
Thermodynamics
148650
A Carnot engine working between $200 \mathrm{~K}$ and $500 \mathrm{~K}$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is
1 $\frac{4000}{3} \mathrm{~J}$
2 $\frac{2000}{3} \mathrm{~J}$
3 $\frac{800}{3} \mathrm{~J}$
4 $\frac{1600}{3} \mathrm{~J}$
Explanation:
A Given that, Initial temperature $\left(\mathrm{T}_{\mathrm{L}}\right)=200 \mathrm{~K}$ Final temperature, $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Work done $=800$ Joule Efficiency, $\eta=\frac{T_{H}-T_{L}}{T_{H}}=\frac{\text { Work done }}{\text { Heat supply }}$ $\frac{500-200}{500}=\frac{800}{\text { Heat supply }}$ $\frac{300}{500}=\frac{800}{\text { Heat supply }}$ 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 $800 \mathrm{~K}$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is
1 $400 \mathrm{~K}$
2 $450 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $550 \mathrm{~K}$
Explanation:
D Given, Initial temperature of $\mathrm{A}=800 \mathrm{~K}$ Final temperature of $B=300 \mathrm{~K}$ Work done for engine $A=$ Work done for engine $B$ $\mathrm{W}_{\mathrm{A}}=\mathrm{W}_{\mathrm{B}}$ So, $\quad 800-\mathrm{T}=\mathrm{T}-300$ $2 \mathrm{~T}=300+800$ $2 \mathrm{~T}=1100$ $\mathrm{T}=\frac{1100}{2}$ $\mathrm{T}=550 \mathrm{~K}$
TS EAMCET (Medical)-02.05.2018
Thermodynamics
148652
A Carnot engine has efficiency of $50 \%$. If the temperature of sink is reduced by $40^{\circ} \mathrm{C}$, its efficiency increases by $30 \%$. The temperature of the source will be:
148647
When the gas expands with temperature using the relation $\mathrm{V}=\mathrm{KT}^{2 / 3}$ for the temperature change of $40 \mathrm{~K}$, the work done is
1 $20.1 \mathrm{R}$
2 $30.2 \mathrm{R}$
3 $26.6 \mathrm{R}$
4 $35.6 \mathrm{R}$
Explanation:
C Given that, $\mathrm{V}=\mathrm{KT}^{2 / 3}$ $\mathrm{dV}=\mathrm{K} \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ Temperature change $(\Delta \mathrm{T})=40 \mathrm{~K}$ Work done, $(\mathrm{dW})=\mathrm{PdV}$ From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\frac{\mathrm{nRT}}{\mathrm{KT}^{\frac{2}{3}}}$ $\mathrm{dW}=\mathrm{P} \cdot \Delta \mathrm{V}$ $\mathrm{dW}=\frac{\mathrm{nRT}}{\mathrm{KT}^{2 / 3}} \cdot K \frac{2}{3} \mathrm{~T}^{\frac{-1}{3}} \mathrm{dT}$ $=\frac{2 \mathrm{nR}}{3}\left(\mathrm{~T}^{1-\frac{1}{3}-\frac{2}{3}}\right) \mathrm{dT}$ $=\frac{2}{3} \mathrm{nR} \int_{0}^{40} \mathrm{dT} \quad(\because \mathrm{n}=1)$ $\therefore \mathrm{dW}=\frac{2}{3} \mathrm{R}[\mathrm{T}]_{0}^{40}$ $=\frac{2}{3} \mathrm{R}(40-0)$ $=\frac{80}{3} \mathrm{R}$ $=26.66 \mathrm{R}$ $\mathrm{dW}=26.67 \mathrm{R}$
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 $62^{\circ} \mathrm{C}$, its efficiency gets doubled. The temperature of the source and sink respectively are.
1 $372 \mathrm{~K}$ and $310 \mathrm{~K}$
2 $273 \mathrm{~K}$ and $300 \mathrm{~K}$
3 $99^{\circ} \mathrm{C}$ and $10^{\circ} \mathrm{C}$
4 $200^{\circ} \mathrm{C}$ and $37^{\circ} \mathrm{C}$
Explanation:
A Given that, Initial efficiency of reversible engine $=\frac{1}{6}$ Decreasing temperature $=62^{\circ} \mathrm{C}$ Final efficiency $=2 \times \frac{1}{6}=\frac{1}{3}$. As we know that, $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ According to question, $\frac{1}{3}=1-\frac{\left(\mathrm{T}_{\mathrm{L}}-62\right)}{\mathrm{T}_{\mathrm{H}}}$ $=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}+62}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{1}{3}=\frac{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}+\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ From equation (i), we get $\frac{1}{3}=\frac{1}{6}+\frac{62}{T_{H}}$ $\frac{1}{3}-\frac{1}{6}=\frac{62}{T_{H}}$ $\frac{1}{6}=\frac{62}{\mathrm{~T}_{\mathrm{H}}}$ $\mathrm{T}_{\mathrm{H}}=372 \mathrm{~K}$ Putting the value of $T_{H}$ in equation (i), we get $\frac{1}{6}=1-\frac{\mathrm{T}_{\mathrm{L}}}{372}$ $\frac{\mathrm{T}_{\mathrm{L}}}{372}=\frac{5}{6}$ $\mathrm{T}_{\mathrm{L}}=\frac{5 \times 372}{6}$ $\mathrm{T}_{\mathrm{L}}=5 \times 62$ $\mathrm{T}_{\mathrm{L}}=310 \mathrm{~K}$
TS EAMCET (Engg.)-2017
Thermodynamics
148650
A Carnot engine working between $200 \mathrm{~K}$ and $500 \mathrm{~K}$ has work done equal to 800 Joules. Amount of heat energy supplied to the engine from the source is
1 $\frac{4000}{3} \mathrm{~J}$
2 $\frac{2000}{3} \mathrm{~J}$
3 $\frac{800}{3} \mathrm{~J}$
4 $\frac{1600}{3} \mathrm{~J}$
Explanation:
A Given that, Initial temperature $\left(\mathrm{T}_{\mathrm{L}}\right)=200 \mathrm{~K}$ Final temperature, $\left(\mathrm{T}_{\mathrm{H}}\right)=500 \mathrm{~K}$ Work done $=800$ Joule Efficiency, $\eta=\frac{T_{H}-T_{L}}{T_{H}}=\frac{\text { Work done }}{\text { Heat supply }}$ $\frac{500-200}{500}=\frac{800}{\text { Heat supply }}$ $\frac{300}{500}=\frac{800}{\text { Heat supply }}$ 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 $800 \mathrm{~K}$ and $T$ and engine $B$ are $T$ and $300 K$ respectively. Then the temperature $T$ is
1 $400 \mathrm{~K}$
2 $450 \mathrm{~K}$
3 $500 \mathrm{~K}$
4 $550 \mathrm{~K}$
Explanation:
D Given, Initial temperature of $\mathrm{A}=800 \mathrm{~K}$ Final temperature of $B=300 \mathrm{~K}$ Work done for engine $A=$ Work done for engine $B$ $\mathrm{W}_{\mathrm{A}}=\mathrm{W}_{\mathrm{B}}$ So, $\quad 800-\mathrm{T}=\mathrm{T}-300$ $2 \mathrm{~T}=300+800$ $2 \mathrm{~T}=1100$ $\mathrm{T}=\frac{1100}{2}$ $\mathrm{T}=550 \mathrm{~K}$
TS EAMCET (Medical)-02.05.2018
Thermodynamics
148652
A Carnot engine has efficiency of $50 \%$. If the temperature of sink is reduced by $40^{\circ} \mathrm{C}$, its efficiency increases by $30 \%$. The temperature of the source will be:
