148581
A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by $62 \mathrm{~K}$, the efficiency of Carnot's cycle becomes $1 / 3$. The temperature of the source and sink will be-
1 $372 \mathrm{~K}, 310 \mathrm{~K}$
2 $181 \mathrm{~K}, 150 \mathrm{~K}$
3 $472 \mathrm{~K}, 410 \mathrm{~K}$
4 none of the above
Explanation:
A Given that, $\eta_{1}=\frac{1}{6}, \quad \eta_{2}=\frac{1}{3}$ We know that $\eta_{1}=\frac{W}{Q}=\frac{1}{6}$ $\eta_{1}=1-\frac{T_{2}}{T_{1}}=\frac{1}{6}$ $\frac{T_{1}-T_{2}}{T_{1}}=\frac{1}{6}$ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}=\mathrm{T}_{1}$ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}-\mathrm{T}_{1}=0$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{5}{6}$ In second case $\frac{1}{3}=\frac{\mathrm{T}_{1}-\left(\mathrm{T}_{2}-62\right)}{\mathrm{T}_{1}}$ $\mathrm{~T}_{1}=3 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}+186$ $2 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}=-186$ Solving equation (i) and (ii), we get $\mathrm{T}_{1} =372 \mathrm{~K}$ $\mathrm{~T}_{2} =\frac{5}{6} \mathrm{~T}_{1}$ $\mathrm{~T}_{2} =\frac{5}{6} \times 372$ $\mathrm{~T}_{2} =310 \mathrm{~K}$ $\text { and } \quad \mathrm{T}_{2}=\frac{5}{6} \mathrm{~T}_{1}$ The temperature of source is $372 \mathrm{~K}$ and sink temperature is $310 \mathrm{~K}$.
BCECE-2007
Thermodynamics
148582
An ideal heat engine exhausting heat at $77^{\circ} \mathrm{C}$ is to have $30 \%$ efficiency. It must take heat at:
1 $127^{\circ} \mathrm{C}$
2 $227^{\circ} \mathrm{C}$
3 $327^{\circ} \mathrm{C}$
4 $673^{\circ} \mathrm{C}$
Explanation:
B Given that, $\eta=30 \%=0.30$ $T_{E}=77^{\circ} \mathrm{C}=77+273=350 \mathrm{~K}$ Heat engine efficiency, $\eta_{\text {H.E. }}=1-\frac{Q_{L}}{Q_{H}}=1-\frac{T_{L}}{T_{H}}$ $0.30=1-\frac{350}{T_{H}}$ $0.30-1=\frac{-350}{T_{H}}$ $-0.70=-\frac{350}{T_{H}}$ $T_{H}=500 \mathrm{~K}$ or $\quad \mathrm{T}_{\mathrm{H}}=227^{\circ} \mathrm{C}$
BCECE-2004
Thermodynamics
148584
A thermodynamic system undergoes a cyclic process $A B C$ as shown in the diagram. The work done by the system per cycle is :
1 $750 \mathrm{~J}$
2 $-1250 \mathrm{~J}$
3 $-750 \mathrm{~J}$
4 $1250 \mathrm{~J}$
Explanation:
C A thermodynamic system undergoes a cyclic process ABC. The work done by the system can be predicted by calculating the total area under the curve. Work done by system $=\frac{1}{2} \times(5-10) \times(400-100)$ $\mathrm{W}=-\frac{1500}{2}$ $\mathrm{~W}=-750 \mathrm{~J}$
Karnataka CET-2019
Thermodynamics
148585
A Carnot engine takes 300 calories of heat from a source at $500 \mathrm{~K}$ and rejects 150 calories of heat to the sink. The temperature of the sink is :
1 $125 \mathrm{~K}$
2 $250 \mathrm{~K}$
3 $750 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
B Given that, $\mathrm{Q}_{1}=300 \mathrm{cal}, \mathrm{Q}_{2}=150 \mathrm{cal} .$ $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=?$ The efficiency of a Carnot engine is, $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{T}_{2}}{500}=\frac{150}{300}$ $\mathrm{~T}_{2}=500 \times \frac{150}{300}$ $\mathrm{~T}_{2}=250 \mathrm{~K}$ Hence, the temperature of the sink is $250 \mathrm{~K}$.
148581
A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by $62 \mathrm{~K}$, the efficiency of Carnot's cycle becomes $1 / 3$. The temperature of the source and sink will be-
1 $372 \mathrm{~K}, 310 \mathrm{~K}$
2 $181 \mathrm{~K}, 150 \mathrm{~K}$
3 $472 \mathrm{~K}, 410 \mathrm{~K}$
4 none of the above
Explanation:
A Given that, $\eta_{1}=\frac{1}{6}, \quad \eta_{2}=\frac{1}{3}$ We know that $\eta_{1}=\frac{W}{Q}=\frac{1}{6}$ $\eta_{1}=1-\frac{T_{2}}{T_{1}}=\frac{1}{6}$ $\frac{T_{1}-T_{2}}{T_{1}}=\frac{1}{6}$ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}=\mathrm{T}_{1}$ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}-\mathrm{T}_{1}=0$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{5}{6}$ In second case $\frac{1}{3}=\frac{\mathrm{T}_{1}-\left(\mathrm{T}_{2}-62\right)}{\mathrm{T}_{1}}$ $\mathrm{~T}_{1}=3 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}+186$ $2 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}=-186$ Solving equation (i) and (ii), we get $\mathrm{T}_{1} =372 \mathrm{~K}$ $\mathrm{~T}_{2} =\frac{5}{6} \mathrm{~T}_{1}$ $\mathrm{~T}_{2} =\frac{5}{6} \times 372$ $\mathrm{~T}_{2} =310 \mathrm{~K}$ $\text { and } \quad \mathrm{T}_{2}=\frac{5}{6} \mathrm{~T}_{1}$ The temperature of source is $372 \mathrm{~K}$ and sink temperature is $310 \mathrm{~K}$.
BCECE-2007
Thermodynamics
148582
An ideal heat engine exhausting heat at $77^{\circ} \mathrm{C}$ is to have $30 \%$ efficiency. It must take heat at:
1 $127^{\circ} \mathrm{C}$
2 $227^{\circ} \mathrm{C}$
3 $327^{\circ} \mathrm{C}$
4 $673^{\circ} \mathrm{C}$
Explanation:
B Given that, $\eta=30 \%=0.30$ $T_{E}=77^{\circ} \mathrm{C}=77+273=350 \mathrm{~K}$ Heat engine efficiency, $\eta_{\text {H.E. }}=1-\frac{Q_{L}}{Q_{H}}=1-\frac{T_{L}}{T_{H}}$ $0.30=1-\frac{350}{T_{H}}$ $0.30-1=\frac{-350}{T_{H}}$ $-0.70=-\frac{350}{T_{H}}$ $T_{H}=500 \mathrm{~K}$ or $\quad \mathrm{T}_{\mathrm{H}}=227^{\circ} \mathrm{C}$
BCECE-2004
Thermodynamics
148584
A thermodynamic system undergoes a cyclic process $A B C$ as shown in the diagram. The work done by the system per cycle is :
1 $750 \mathrm{~J}$
2 $-1250 \mathrm{~J}$
3 $-750 \mathrm{~J}$
4 $1250 \mathrm{~J}$
Explanation:
C A thermodynamic system undergoes a cyclic process ABC. The work done by the system can be predicted by calculating the total area under the curve. Work done by system $=\frac{1}{2} \times(5-10) \times(400-100)$ $\mathrm{W}=-\frac{1500}{2}$ $\mathrm{~W}=-750 \mathrm{~J}$
Karnataka CET-2019
Thermodynamics
148585
A Carnot engine takes 300 calories of heat from a source at $500 \mathrm{~K}$ and rejects 150 calories of heat to the sink. The temperature of the sink is :
1 $125 \mathrm{~K}$
2 $250 \mathrm{~K}$
3 $750 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
B Given that, $\mathrm{Q}_{1}=300 \mathrm{cal}, \mathrm{Q}_{2}=150 \mathrm{cal} .$ $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=?$ The efficiency of a Carnot engine is, $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{T}_{2}}{500}=\frac{150}{300}$ $\mathrm{~T}_{2}=500 \times \frac{150}{300}$ $\mathrm{~T}_{2}=250 \mathrm{~K}$ Hence, the temperature of the sink is $250 \mathrm{~K}$.
148581
A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by $62 \mathrm{~K}$, the efficiency of Carnot's cycle becomes $1 / 3$. The temperature of the source and sink will be-
1 $372 \mathrm{~K}, 310 \mathrm{~K}$
2 $181 \mathrm{~K}, 150 \mathrm{~K}$
3 $472 \mathrm{~K}, 410 \mathrm{~K}$
4 none of the above
Explanation:
A Given that, $\eta_{1}=\frac{1}{6}, \quad \eta_{2}=\frac{1}{3}$ We know that $\eta_{1}=\frac{W}{Q}=\frac{1}{6}$ $\eta_{1}=1-\frac{T_{2}}{T_{1}}=\frac{1}{6}$ $\frac{T_{1}-T_{2}}{T_{1}}=\frac{1}{6}$ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}=\mathrm{T}_{1}$ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}-\mathrm{T}_{1}=0$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{5}{6}$ In second case $\frac{1}{3}=\frac{\mathrm{T}_{1}-\left(\mathrm{T}_{2}-62\right)}{\mathrm{T}_{1}}$ $\mathrm{~T}_{1}=3 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}+186$ $2 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}=-186$ Solving equation (i) and (ii), we get $\mathrm{T}_{1} =372 \mathrm{~K}$ $\mathrm{~T}_{2} =\frac{5}{6} \mathrm{~T}_{1}$ $\mathrm{~T}_{2} =\frac{5}{6} \times 372$ $\mathrm{~T}_{2} =310 \mathrm{~K}$ $\text { and } \quad \mathrm{T}_{2}=\frac{5}{6} \mathrm{~T}_{1}$ The temperature of source is $372 \mathrm{~K}$ and sink temperature is $310 \mathrm{~K}$.
BCECE-2007
Thermodynamics
148582
An ideal heat engine exhausting heat at $77^{\circ} \mathrm{C}$ is to have $30 \%$ efficiency. It must take heat at:
1 $127^{\circ} \mathrm{C}$
2 $227^{\circ} \mathrm{C}$
3 $327^{\circ} \mathrm{C}$
4 $673^{\circ} \mathrm{C}$
Explanation:
B Given that, $\eta=30 \%=0.30$ $T_{E}=77^{\circ} \mathrm{C}=77+273=350 \mathrm{~K}$ Heat engine efficiency, $\eta_{\text {H.E. }}=1-\frac{Q_{L}}{Q_{H}}=1-\frac{T_{L}}{T_{H}}$ $0.30=1-\frac{350}{T_{H}}$ $0.30-1=\frac{-350}{T_{H}}$ $-0.70=-\frac{350}{T_{H}}$ $T_{H}=500 \mathrm{~K}$ or $\quad \mathrm{T}_{\mathrm{H}}=227^{\circ} \mathrm{C}$
BCECE-2004
Thermodynamics
148584
A thermodynamic system undergoes a cyclic process $A B C$ as shown in the diagram. The work done by the system per cycle is :
1 $750 \mathrm{~J}$
2 $-1250 \mathrm{~J}$
3 $-750 \mathrm{~J}$
4 $1250 \mathrm{~J}$
Explanation:
C A thermodynamic system undergoes a cyclic process ABC. The work done by the system can be predicted by calculating the total area under the curve. Work done by system $=\frac{1}{2} \times(5-10) \times(400-100)$ $\mathrm{W}=-\frac{1500}{2}$ $\mathrm{~W}=-750 \mathrm{~J}$
Karnataka CET-2019
Thermodynamics
148585
A Carnot engine takes 300 calories of heat from a source at $500 \mathrm{~K}$ and rejects 150 calories of heat to the sink. The temperature of the sink is :
1 $125 \mathrm{~K}$
2 $250 \mathrm{~K}$
3 $750 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
B Given that, $\mathrm{Q}_{1}=300 \mathrm{cal}, \mathrm{Q}_{2}=150 \mathrm{cal} .$ $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=?$ The efficiency of a Carnot engine is, $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{T}_{2}}{500}=\frac{150}{300}$ $\mathrm{~T}_{2}=500 \times \frac{150}{300}$ $\mathrm{~T}_{2}=250 \mathrm{~K}$ Hence, the temperature of the sink is $250 \mathrm{~K}$.
148581
A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by $62 \mathrm{~K}$, the efficiency of Carnot's cycle becomes $1 / 3$. The temperature of the source and sink will be-
1 $372 \mathrm{~K}, 310 \mathrm{~K}$
2 $181 \mathrm{~K}, 150 \mathrm{~K}$
3 $472 \mathrm{~K}, 410 \mathrm{~K}$
4 none of the above
Explanation:
A Given that, $\eta_{1}=\frac{1}{6}, \quad \eta_{2}=\frac{1}{3}$ We know that $\eta_{1}=\frac{W}{Q}=\frac{1}{6}$ $\eta_{1}=1-\frac{T_{2}}{T_{1}}=\frac{1}{6}$ $\frac{T_{1}-T_{2}}{T_{1}}=\frac{1}{6}$ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}=\mathrm{T}_{1}$ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}-\mathrm{T}_{1}=0$ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{5}{6}$ In second case $\frac{1}{3}=\frac{\mathrm{T}_{1}-\left(\mathrm{T}_{2}-62\right)}{\mathrm{T}_{1}}$ $\mathrm{~T}_{1}=3 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}+186$ $2 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}=-186$ Solving equation (i) and (ii), we get $\mathrm{T}_{1} =372 \mathrm{~K}$ $\mathrm{~T}_{2} =\frac{5}{6} \mathrm{~T}_{1}$ $\mathrm{~T}_{2} =\frac{5}{6} \times 372$ $\mathrm{~T}_{2} =310 \mathrm{~K}$ $\text { and } \quad \mathrm{T}_{2}=\frac{5}{6} \mathrm{~T}_{1}$ The temperature of source is $372 \mathrm{~K}$ and sink temperature is $310 \mathrm{~K}$.
BCECE-2007
Thermodynamics
148582
An ideal heat engine exhausting heat at $77^{\circ} \mathrm{C}$ is to have $30 \%$ efficiency. It must take heat at:
1 $127^{\circ} \mathrm{C}$
2 $227^{\circ} \mathrm{C}$
3 $327^{\circ} \mathrm{C}$
4 $673^{\circ} \mathrm{C}$
Explanation:
B Given that, $\eta=30 \%=0.30$ $T_{E}=77^{\circ} \mathrm{C}=77+273=350 \mathrm{~K}$ Heat engine efficiency, $\eta_{\text {H.E. }}=1-\frac{Q_{L}}{Q_{H}}=1-\frac{T_{L}}{T_{H}}$ $0.30=1-\frac{350}{T_{H}}$ $0.30-1=\frac{-350}{T_{H}}$ $-0.70=-\frac{350}{T_{H}}$ $T_{H}=500 \mathrm{~K}$ or $\quad \mathrm{T}_{\mathrm{H}}=227^{\circ} \mathrm{C}$
BCECE-2004
Thermodynamics
148584
A thermodynamic system undergoes a cyclic process $A B C$ as shown in the diagram. The work done by the system per cycle is :
1 $750 \mathrm{~J}$
2 $-1250 \mathrm{~J}$
3 $-750 \mathrm{~J}$
4 $1250 \mathrm{~J}$
Explanation:
C A thermodynamic system undergoes a cyclic process ABC. The work done by the system can be predicted by calculating the total area under the curve. Work done by system $=\frac{1}{2} \times(5-10) \times(400-100)$ $\mathrm{W}=-\frac{1500}{2}$ $\mathrm{~W}=-750 \mathrm{~J}$
Karnataka CET-2019
Thermodynamics
148585
A Carnot engine takes 300 calories of heat from a source at $500 \mathrm{~K}$ and rejects 150 calories of heat to the sink. The temperature of the sink is :
1 $125 \mathrm{~K}$
2 $250 \mathrm{~K}$
3 $750 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
B Given that, $\mathrm{Q}_{1}=300 \mathrm{cal}, \mathrm{Q}_{2}=150 \mathrm{cal} .$ $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=?$ The efficiency of a Carnot engine is, $1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$ $\frac{\mathrm{T}_{2}}{500}=\frac{150}{300}$ $\mathrm{~T}_{2}=500 \times \frac{150}{300}$ $\mathrm{~T}_{2}=250 \mathrm{~K}$ Hence, the temperature of the sink is $250 \mathrm{~K}$.
