148602
The efficiency of a Carnot engine is $60 \%$. If the temperature of the sink is $27^{\circ} \mathrm{C}$, the temperature of the source is
1 $187.5^{\circ} \mathrm{C}$
2 $207^{\circ} \mathrm{C}$
3 $477^{\circ} \mathrm{C}$
4 $750^{\circ} \mathrm{C}$
Explanation:
C Given that Efficiency of engine $=60 \%$ Sink Temperature, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273$ Source Temperature, $\mathrm{T}_{1}=$ ? As we know that, Efficiency of carnot engine $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ $\eta=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $\frac{60}{100}=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $\frac{3}{5}=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $5\left(\mathrm{~T}_{1}-300\right)=3 \mathrm{~T}_{1}$ $5 \mathrm{~T}_{1}-1500=3 \mathrm{~T}_{1}$ $2 \mathrm{~T}_{1}=1500$ $\mathrm{T}_{1} =\frac{1500}{2}$ $\mathrm{~T}_{1} =750 \mathrm{~K}=(750-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{1} =477^{\circ} \mathrm{C}$
J and K CET- 1998
Thermodynamics
148603
Calculate the efficiency of the engine if Carnot cycle operates at $T_{1}=550 \mathrm{~K}$ and $T_{2}=320 \mathrm{~K}$ producing $2.3 \mathrm{~kJ}$ of mechanical work per cycle?
1 0.418
2 0.622
3 0.823
4 0.902
Explanation:
A Given that, $\mathrm{T}_{1}=550 \mathrm{~K}$ $\mathrm{T}_{2}=320 \mathrm{~K}$ Efficiency of carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ $=1-\frac{320}{550}$ $=1-0.582$ $\eta =0.418$
J and K-CET-2014
Thermodynamics
148604
The efficiency of a Carnot engine kept at the temperatures of $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ is
1 $20 \%$
2 $25 \%$
3 $30 \%$
4 $40 \%$
Explanation:
B Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ We know that, Efficiency of a Carnot engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $\eta =\left(1-\frac{300}{400}\right) \times 100$ $\eta =0.25 \times 100$ $\eta =25 \%$
J and K-CET-2013
Thermodynamics
148605
An ideal Carnot's engine works between $227^{\circ} \mathrm{C}$ and $57^{\circ} \mathrm{C}$. Find the efficiency of the engine
1 $22 \%$
2 $34 \%$
3 $55 \%$
4 $13.5 \%$
Explanation:
B Given that, $\mathrm{T}_{1}=227^{\circ} \mathrm{C}=(227+273) \mathrm{K}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=57^{\circ} \mathrm{C}=(57+273) \mathrm{K}=330 \mathrm{~K}$ Efficiency of Carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{330}{500}=0.34$ $\eta=34 \%$
148602
The efficiency of a Carnot engine is $60 \%$. If the temperature of the sink is $27^{\circ} \mathrm{C}$, the temperature of the source is
1 $187.5^{\circ} \mathrm{C}$
2 $207^{\circ} \mathrm{C}$
3 $477^{\circ} \mathrm{C}$
4 $750^{\circ} \mathrm{C}$
Explanation:
C Given that Efficiency of engine $=60 \%$ Sink Temperature, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273$ Source Temperature, $\mathrm{T}_{1}=$ ? As we know that, Efficiency of carnot engine $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ $\eta=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $\frac{60}{100}=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $\frac{3}{5}=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $5\left(\mathrm{~T}_{1}-300\right)=3 \mathrm{~T}_{1}$ $5 \mathrm{~T}_{1}-1500=3 \mathrm{~T}_{1}$ $2 \mathrm{~T}_{1}=1500$ $\mathrm{T}_{1} =\frac{1500}{2}$ $\mathrm{~T}_{1} =750 \mathrm{~K}=(750-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{1} =477^{\circ} \mathrm{C}$
J and K CET- 1998
Thermodynamics
148603
Calculate the efficiency of the engine if Carnot cycle operates at $T_{1}=550 \mathrm{~K}$ and $T_{2}=320 \mathrm{~K}$ producing $2.3 \mathrm{~kJ}$ of mechanical work per cycle?
1 0.418
2 0.622
3 0.823
4 0.902
Explanation:
A Given that, $\mathrm{T}_{1}=550 \mathrm{~K}$ $\mathrm{T}_{2}=320 \mathrm{~K}$ Efficiency of carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ $=1-\frac{320}{550}$ $=1-0.582$ $\eta =0.418$
J and K-CET-2014
Thermodynamics
148604
The efficiency of a Carnot engine kept at the temperatures of $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ is
1 $20 \%$
2 $25 \%$
3 $30 \%$
4 $40 \%$
Explanation:
B Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ We know that, Efficiency of a Carnot engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $\eta =\left(1-\frac{300}{400}\right) \times 100$ $\eta =0.25 \times 100$ $\eta =25 \%$
J and K-CET-2013
Thermodynamics
148605
An ideal Carnot's engine works between $227^{\circ} \mathrm{C}$ and $57^{\circ} \mathrm{C}$. Find the efficiency of the engine
1 $22 \%$
2 $34 \%$
3 $55 \%$
4 $13.5 \%$
Explanation:
B Given that, $\mathrm{T}_{1}=227^{\circ} \mathrm{C}=(227+273) \mathrm{K}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=57^{\circ} \mathrm{C}=(57+273) \mathrm{K}=330 \mathrm{~K}$ Efficiency of Carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{330}{500}=0.34$ $\eta=34 \%$
148602
The efficiency of a Carnot engine is $60 \%$. If the temperature of the sink is $27^{\circ} \mathrm{C}$, the temperature of the source is
1 $187.5^{\circ} \mathrm{C}$
2 $207^{\circ} \mathrm{C}$
3 $477^{\circ} \mathrm{C}$
4 $750^{\circ} \mathrm{C}$
Explanation:
C Given that Efficiency of engine $=60 \%$ Sink Temperature, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273$ Source Temperature, $\mathrm{T}_{1}=$ ? As we know that, Efficiency of carnot engine $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ $\eta=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $\frac{60}{100}=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $\frac{3}{5}=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $5\left(\mathrm{~T}_{1}-300\right)=3 \mathrm{~T}_{1}$ $5 \mathrm{~T}_{1}-1500=3 \mathrm{~T}_{1}$ $2 \mathrm{~T}_{1}=1500$ $\mathrm{T}_{1} =\frac{1500}{2}$ $\mathrm{~T}_{1} =750 \mathrm{~K}=(750-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{1} =477^{\circ} \mathrm{C}$
J and K CET- 1998
Thermodynamics
148603
Calculate the efficiency of the engine if Carnot cycle operates at $T_{1}=550 \mathrm{~K}$ and $T_{2}=320 \mathrm{~K}$ producing $2.3 \mathrm{~kJ}$ of mechanical work per cycle?
1 0.418
2 0.622
3 0.823
4 0.902
Explanation:
A Given that, $\mathrm{T}_{1}=550 \mathrm{~K}$ $\mathrm{T}_{2}=320 \mathrm{~K}$ Efficiency of carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ $=1-\frac{320}{550}$ $=1-0.582$ $\eta =0.418$
J and K-CET-2014
Thermodynamics
148604
The efficiency of a Carnot engine kept at the temperatures of $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ is
1 $20 \%$
2 $25 \%$
3 $30 \%$
4 $40 \%$
Explanation:
B Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ We know that, Efficiency of a Carnot engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $\eta =\left(1-\frac{300}{400}\right) \times 100$ $\eta =0.25 \times 100$ $\eta =25 \%$
J and K-CET-2013
Thermodynamics
148605
An ideal Carnot's engine works between $227^{\circ} \mathrm{C}$ and $57^{\circ} \mathrm{C}$. Find the efficiency of the engine
1 $22 \%$
2 $34 \%$
3 $55 \%$
4 $13.5 \%$
Explanation:
B Given that, $\mathrm{T}_{1}=227^{\circ} \mathrm{C}=(227+273) \mathrm{K}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=57^{\circ} \mathrm{C}=(57+273) \mathrm{K}=330 \mathrm{~K}$ Efficiency of Carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{330}{500}=0.34$ $\eta=34 \%$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
148602
The efficiency of a Carnot engine is $60 \%$. If the temperature of the sink is $27^{\circ} \mathrm{C}$, the temperature of the source is
1 $187.5^{\circ} \mathrm{C}$
2 $207^{\circ} \mathrm{C}$
3 $477^{\circ} \mathrm{C}$
4 $750^{\circ} \mathrm{C}$
Explanation:
C Given that Efficiency of engine $=60 \%$ Sink Temperature, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273$ Source Temperature, $\mathrm{T}_{1}=$ ? As we know that, Efficiency of carnot engine $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ $\eta=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $\frac{60}{100}=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $\frac{3}{5}=\frac{\mathrm{T}_{1}-300}{\mathrm{~T}_{1}}$ $5\left(\mathrm{~T}_{1}-300\right)=3 \mathrm{~T}_{1}$ $5 \mathrm{~T}_{1}-1500=3 \mathrm{~T}_{1}$ $2 \mathrm{~T}_{1}=1500$ $\mathrm{T}_{1} =\frac{1500}{2}$ $\mathrm{~T}_{1} =750 \mathrm{~K}=(750-273)^{\circ} \mathrm{C}$ $\mathrm{T}_{1} =477^{\circ} \mathrm{C}$
J and K CET- 1998
Thermodynamics
148603
Calculate the efficiency of the engine if Carnot cycle operates at $T_{1}=550 \mathrm{~K}$ and $T_{2}=320 \mathrm{~K}$ producing $2.3 \mathrm{~kJ}$ of mechanical work per cycle?
1 0.418
2 0.622
3 0.823
4 0.902
Explanation:
A Given that, $\mathrm{T}_{1}=550 \mathrm{~K}$ $\mathrm{T}_{2}=320 \mathrm{~K}$ Efficiency of carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ $=1-\frac{320}{550}$ $=1-0.582$ $\eta =0.418$
J and K-CET-2014
Thermodynamics
148604
The efficiency of a Carnot engine kept at the temperatures of $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ is
1 $20 \%$
2 $25 \%$
3 $30 \%$
4 $40 \%$
Explanation:
B Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ We know that, Efficiency of a Carnot engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $\eta =\left(1-\frac{300}{400}\right) \times 100$ $\eta =0.25 \times 100$ $\eta =25 \%$
J and K-CET-2013
Thermodynamics
148605
An ideal Carnot's engine works between $227^{\circ} \mathrm{C}$ and $57^{\circ} \mathrm{C}$. Find the efficiency of the engine
1 $22 \%$
2 $34 \%$
3 $55 \%$
4 $13.5 \%$
Explanation:
B Given that, $\mathrm{T}_{1}=227^{\circ} \mathrm{C}=(227+273) \mathrm{K}=500 \mathrm{~K}$ $\mathrm{~T}_{2}=57^{\circ} \mathrm{C}=(57+273) \mathrm{K}=330 \mathrm{~K}$ Efficiency of Carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{330}{500}=0.34$ $\eta=34 \%$
