148634
A Carnot engine whose efficiency is $40 \%$ takes in heat from source maintained at a temperature of $500 \mathrm{~K}$. It is desired to have an engine of efficiency $60 \%$. Then, the intake temperature for the same exhaust (sink) temperature must be
1 $1200 \mathrm{~K}$
2 $750 \mathrm{~K}$
3 $600 \mathrm{~K}$
4 $800 \mathrm{~K}$
Explanation:
B For Carnot engine, the efficiency is given by $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $\mathrm{T}_{\mathrm{L}}=$ Low temperature (sink) $\mathrm{T}_{\mathrm{H}}=$ High temperature $($ source) $=500 \mathrm{~K}$ Efficiency is given as $40 \%$ $\frac{40}{100}=1-\frac{T_{L}}{500}$ $\frac{T_{L}}{500}=\frac{60}{100}$ $\mathrm{~T}_{\mathrm{L}}=300 \mathrm{~K}$ For efficiency $60 \%$ $\frac{60}{100}=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{300}{\mathrm{~T}_{\mathrm{H}}}=1-\frac{60}{100}$ $\frac{300}{\mathrm{~T}_{\mathrm{H}}}=\frac{40}{100}$ $\mathrm{~T}_{\mathrm{H}}=\frac{300 \times 100}{40}$ $\mathrm{~T}_{\mathrm{H}}=750 \mathrm{~K}$
AP EAMCET-25.08.2021
Thermodynamics
148635
A Carnot engine having an efficiency $1 / 5$ as a heat engine, is used as a refrigerator. If the work done on the system is $50 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is
1 $90 \mathrm{~J}$
2 $99 \mathrm{~J}$
3 $200 \mathrm{~J}$
4 $1 \mathrm{~J}$
Explanation:
C Given, $\eta=\frac{1}{5}=\frac{\text { Work done }}{\text { Heat input }}$ Work done $=50$ Joule We know, the efficiency of Carnot engine $\eta=\frac{Q_{H}-Q_{L}}{Q_{H}}$ Coefficient of performance of a refrigerator, $\alpha =\frac{1-\eta}{\eta}$ $\alpha=\frac{1-\frac{1}{5}}{1 / 5}=\frac{4}{5} \times \frac{5}{1}$ $\alpha=4$ $\text { Also, } \alpha=\frac{Q_{L}}{W} \text { (Where, W is the work done) }$ $Q_{L} =\alpha \times W$ $Q_{L} =4 \times 50$ $Q_{L} =200 \text { Joule }$
AP EAMCET-25.08.2021
Thermodynamics
148636
Three designs are proposed for an engine which is to operate between $500 \mathrm{~K}$ and $300 \mathrm{~K}$. Design A claims to produce $150 \mathrm{~J}$ of work per $1000 \mathrm{~J}$ of heat input, design $B 450 \mathrm{~J}$ of work per $1000 \mathrm{~J}$ and design $\mathrm{C} 300 \mathrm{~J}$ of work per $1000 \mathrm{~J}$. Which of the designs would you choose?
1 $\mathrm{A}$
2 B
3 $\mathrm{C}$
4 None is suitable
Explanation:
C Given that, $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ Efficiency of the engine $(\eta)=1-\frac{T_{L}}{T_{H}}$ $=1-\frac{300}{500}=\frac{2}{5}=0.4$ For design A, Work done $=150 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{A}} =\frac{\text { Work done }}{\text { Heat input }}$ $\eta_{\mathrm{A}} =\frac{150}{1000}=0.15$ For design B, Work done $=450 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{B}}=\frac{450}{1000}=0.45$ For design $\mathrm{C}$, Work done $=300 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{C}}=\frac{300}{1000}=0.3$ Design B has maximum efficiency and our priority is to select engine with maximum efficiency but since $B$ is an impossible engine so the next engine to be selected is C. Note: Official answer given by commission is (b).
AP EAMCET-25.08.2021
Thermodynamics
148637
Three Carnot engines operate in series between a heat source at temperature $T_{1}$ and heat sink at a temperature $T_{4}$. There are two other reservoirs at temperatures $T_{2}$ and $T_{3}$. The three engines are equally efficient if (given that $T_{1}>T_{2}>T_{3}>T_{4}$ )
148634
A Carnot engine whose efficiency is $40 \%$ takes in heat from source maintained at a temperature of $500 \mathrm{~K}$. It is desired to have an engine of efficiency $60 \%$. Then, the intake temperature for the same exhaust (sink) temperature must be
1 $1200 \mathrm{~K}$
2 $750 \mathrm{~K}$
3 $600 \mathrm{~K}$
4 $800 \mathrm{~K}$
Explanation:
B For Carnot engine, the efficiency is given by $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $\mathrm{T}_{\mathrm{L}}=$ Low temperature (sink) $\mathrm{T}_{\mathrm{H}}=$ High temperature $($ source) $=500 \mathrm{~K}$ Efficiency is given as $40 \%$ $\frac{40}{100}=1-\frac{T_{L}}{500}$ $\frac{T_{L}}{500}=\frac{60}{100}$ $\mathrm{~T}_{\mathrm{L}}=300 \mathrm{~K}$ For efficiency $60 \%$ $\frac{60}{100}=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{300}{\mathrm{~T}_{\mathrm{H}}}=1-\frac{60}{100}$ $\frac{300}{\mathrm{~T}_{\mathrm{H}}}=\frac{40}{100}$ $\mathrm{~T}_{\mathrm{H}}=\frac{300 \times 100}{40}$ $\mathrm{~T}_{\mathrm{H}}=750 \mathrm{~K}$
AP EAMCET-25.08.2021
Thermodynamics
148635
A Carnot engine having an efficiency $1 / 5$ as a heat engine, is used as a refrigerator. If the work done on the system is $50 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is
1 $90 \mathrm{~J}$
2 $99 \mathrm{~J}$
3 $200 \mathrm{~J}$
4 $1 \mathrm{~J}$
Explanation:
C Given, $\eta=\frac{1}{5}=\frac{\text { Work done }}{\text { Heat input }}$ Work done $=50$ Joule We know, the efficiency of Carnot engine $\eta=\frac{Q_{H}-Q_{L}}{Q_{H}}$ Coefficient of performance of a refrigerator, $\alpha =\frac{1-\eta}{\eta}$ $\alpha=\frac{1-\frac{1}{5}}{1 / 5}=\frac{4}{5} \times \frac{5}{1}$ $\alpha=4$ $\text { Also, } \alpha=\frac{Q_{L}}{W} \text { (Where, W is the work done) }$ $Q_{L} =\alpha \times W$ $Q_{L} =4 \times 50$ $Q_{L} =200 \text { Joule }$
AP EAMCET-25.08.2021
Thermodynamics
148636
Three designs are proposed for an engine which is to operate between $500 \mathrm{~K}$ and $300 \mathrm{~K}$. Design A claims to produce $150 \mathrm{~J}$ of work per $1000 \mathrm{~J}$ of heat input, design $B 450 \mathrm{~J}$ of work per $1000 \mathrm{~J}$ and design $\mathrm{C} 300 \mathrm{~J}$ of work per $1000 \mathrm{~J}$. Which of the designs would you choose?
1 $\mathrm{A}$
2 B
3 $\mathrm{C}$
4 None is suitable
Explanation:
C Given that, $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ Efficiency of the engine $(\eta)=1-\frac{T_{L}}{T_{H}}$ $=1-\frac{300}{500}=\frac{2}{5}=0.4$ For design A, Work done $=150 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{A}} =\frac{\text { Work done }}{\text { Heat input }}$ $\eta_{\mathrm{A}} =\frac{150}{1000}=0.15$ For design B, Work done $=450 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{B}}=\frac{450}{1000}=0.45$ For design $\mathrm{C}$, Work done $=300 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{C}}=\frac{300}{1000}=0.3$ Design B has maximum efficiency and our priority is to select engine with maximum efficiency but since $B$ is an impossible engine so the next engine to be selected is C. Note: Official answer given by commission is (b).
AP EAMCET-25.08.2021
Thermodynamics
148637
Three Carnot engines operate in series between a heat source at temperature $T_{1}$ and heat sink at a temperature $T_{4}$. There are two other reservoirs at temperatures $T_{2}$ and $T_{3}$. The three engines are equally efficient if (given that $T_{1}>T_{2}>T_{3}>T_{4}$ )
148634
A Carnot engine whose efficiency is $40 \%$ takes in heat from source maintained at a temperature of $500 \mathrm{~K}$. It is desired to have an engine of efficiency $60 \%$. Then, the intake temperature for the same exhaust (sink) temperature must be
1 $1200 \mathrm{~K}$
2 $750 \mathrm{~K}$
3 $600 \mathrm{~K}$
4 $800 \mathrm{~K}$
Explanation:
B For Carnot engine, the efficiency is given by $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $\mathrm{T}_{\mathrm{L}}=$ Low temperature (sink) $\mathrm{T}_{\mathrm{H}}=$ High temperature $($ source) $=500 \mathrm{~K}$ Efficiency is given as $40 \%$ $\frac{40}{100}=1-\frac{T_{L}}{500}$ $\frac{T_{L}}{500}=\frac{60}{100}$ $\mathrm{~T}_{\mathrm{L}}=300 \mathrm{~K}$ For efficiency $60 \%$ $\frac{60}{100}=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{300}{\mathrm{~T}_{\mathrm{H}}}=1-\frac{60}{100}$ $\frac{300}{\mathrm{~T}_{\mathrm{H}}}=\frac{40}{100}$ $\mathrm{~T}_{\mathrm{H}}=\frac{300 \times 100}{40}$ $\mathrm{~T}_{\mathrm{H}}=750 \mathrm{~K}$
AP EAMCET-25.08.2021
Thermodynamics
148635
A Carnot engine having an efficiency $1 / 5$ as a heat engine, is used as a refrigerator. If the work done on the system is $50 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is
1 $90 \mathrm{~J}$
2 $99 \mathrm{~J}$
3 $200 \mathrm{~J}$
4 $1 \mathrm{~J}$
Explanation:
C Given, $\eta=\frac{1}{5}=\frac{\text { Work done }}{\text { Heat input }}$ Work done $=50$ Joule We know, the efficiency of Carnot engine $\eta=\frac{Q_{H}-Q_{L}}{Q_{H}}$ Coefficient of performance of a refrigerator, $\alpha =\frac{1-\eta}{\eta}$ $\alpha=\frac{1-\frac{1}{5}}{1 / 5}=\frac{4}{5} \times \frac{5}{1}$ $\alpha=4$ $\text { Also, } \alpha=\frac{Q_{L}}{W} \text { (Where, W is the work done) }$ $Q_{L} =\alpha \times W$ $Q_{L} =4 \times 50$ $Q_{L} =200 \text { Joule }$
AP EAMCET-25.08.2021
Thermodynamics
148636
Three designs are proposed for an engine which is to operate between $500 \mathrm{~K}$ and $300 \mathrm{~K}$. Design A claims to produce $150 \mathrm{~J}$ of work per $1000 \mathrm{~J}$ of heat input, design $B 450 \mathrm{~J}$ of work per $1000 \mathrm{~J}$ and design $\mathrm{C} 300 \mathrm{~J}$ of work per $1000 \mathrm{~J}$. Which of the designs would you choose?
1 $\mathrm{A}$
2 B
3 $\mathrm{C}$
4 None is suitable
Explanation:
C Given that, $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ Efficiency of the engine $(\eta)=1-\frac{T_{L}}{T_{H}}$ $=1-\frac{300}{500}=\frac{2}{5}=0.4$ For design A, Work done $=150 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{A}} =\frac{\text { Work done }}{\text { Heat input }}$ $\eta_{\mathrm{A}} =\frac{150}{1000}=0.15$ For design B, Work done $=450 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{B}}=\frac{450}{1000}=0.45$ For design $\mathrm{C}$, Work done $=300 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{C}}=\frac{300}{1000}=0.3$ Design B has maximum efficiency and our priority is to select engine with maximum efficiency but since $B$ is an impossible engine so the next engine to be selected is C. Note: Official answer given by commission is (b).
AP EAMCET-25.08.2021
Thermodynamics
148637
Three Carnot engines operate in series between a heat source at temperature $T_{1}$ and heat sink at a temperature $T_{4}$. There are two other reservoirs at temperatures $T_{2}$ and $T_{3}$. The three engines are equally efficient if (given that $T_{1}>T_{2}>T_{3}>T_{4}$ )
148634
A Carnot engine whose efficiency is $40 \%$ takes in heat from source maintained at a temperature of $500 \mathrm{~K}$. It is desired to have an engine of efficiency $60 \%$. Then, the intake temperature for the same exhaust (sink) temperature must be
1 $1200 \mathrm{~K}$
2 $750 \mathrm{~K}$
3 $600 \mathrm{~K}$
4 $800 \mathrm{~K}$
Explanation:
B For Carnot engine, the efficiency is given by $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $\mathrm{T}_{\mathrm{L}}=$ Low temperature (sink) $\mathrm{T}_{\mathrm{H}}=$ High temperature $($ source) $=500 \mathrm{~K}$ Efficiency is given as $40 \%$ $\frac{40}{100}=1-\frac{T_{L}}{500}$ $\frac{T_{L}}{500}=\frac{60}{100}$ $\mathrm{~T}_{\mathrm{L}}=300 \mathrm{~K}$ For efficiency $60 \%$ $\frac{60}{100}=1-\frac{300}{\mathrm{~T}_{\mathrm{H}}}$ $\frac{300}{\mathrm{~T}_{\mathrm{H}}}=1-\frac{60}{100}$ $\frac{300}{\mathrm{~T}_{\mathrm{H}}}=\frac{40}{100}$ $\mathrm{~T}_{\mathrm{H}}=\frac{300 \times 100}{40}$ $\mathrm{~T}_{\mathrm{H}}=750 \mathrm{~K}$
AP EAMCET-25.08.2021
Thermodynamics
148635
A Carnot engine having an efficiency $1 / 5$ as a heat engine, is used as a refrigerator. If the work done on the system is $50 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is
1 $90 \mathrm{~J}$
2 $99 \mathrm{~J}$
3 $200 \mathrm{~J}$
4 $1 \mathrm{~J}$
Explanation:
C Given, $\eta=\frac{1}{5}=\frac{\text { Work done }}{\text { Heat input }}$ Work done $=50$ Joule We know, the efficiency of Carnot engine $\eta=\frac{Q_{H}-Q_{L}}{Q_{H}}$ Coefficient of performance of a refrigerator, $\alpha =\frac{1-\eta}{\eta}$ $\alpha=\frac{1-\frac{1}{5}}{1 / 5}=\frac{4}{5} \times \frac{5}{1}$ $\alpha=4$ $\text { Also, } \alpha=\frac{Q_{L}}{W} \text { (Where, W is the work done) }$ $Q_{L} =\alpha \times W$ $Q_{L} =4 \times 50$ $Q_{L} =200 \text { Joule }$
AP EAMCET-25.08.2021
Thermodynamics
148636
Three designs are proposed for an engine which is to operate between $500 \mathrm{~K}$ and $300 \mathrm{~K}$. Design A claims to produce $150 \mathrm{~J}$ of work per $1000 \mathrm{~J}$ of heat input, design $B 450 \mathrm{~J}$ of work per $1000 \mathrm{~J}$ and design $\mathrm{C} 300 \mathrm{~J}$ of work per $1000 \mathrm{~J}$. Which of the designs would you choose?
1 $\mathrm{A}$
2 B
3 $\mathrm{C}$
4 None is suitable
Explanation:
C Given that, $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ Efficiency of the engine $(\eta)=1-\frac{T_{L}}{T_{H}}$ $=1-\frac{300}{500}=\frac{2}{5}=0.4$ For design A, Work done $=150 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{A}} =\frac{\text { Work done }}{\text { Heat input }}$ $\eta_{\mathrm{A}} =\frac{150}{1000}=0.15$ For design B, Work done $=450 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{B}}=\frac{450}{1000}=0.45$ For design $\mathrm{C}$, Work done $=300 \mathrm{~J}$, Heat input $=1000 \mathrm{~J}$ $\eta_{\mathrm{C}}=\frac{300}{1000}=0.3$ Design B has maximum efficiency and our priority is to select engine with maximum efficiency but since $B$ is an impossible engine so the next engine to be selected is C. Note: Official answer given by commission is (b).
AP EAMCET-25.08.2021
Thermodynamics
148637
Three Carnot engines operate in series between a heat source at temperature $T_{1}$ and heat sink at a temperature $T_{4}$. There are two other reservoirs at temperatures $T_{2}$ and $T_{3}$. The three engines are equally efficient if (given that $T_{1}>T_{2}>T_{3}>T_{4}$ )
