148586
For which combination of working temperatures, the efficiency of Carnot's engine is the least ?
1 $100 \mathrm{~K}, 80 \mathrm{~K}$
2 $40 \mathrm{~K}, 20 \mathrm{~K}$
3 $80 \mathrm{~K}, 60 \mathrm{~K}$
4 $60 \mathrm{~K}, 40 \mathrm{~K}$
Explanation:
A We know that the efficiency of Carnot's Engine $(\eta)=1-\frac{T_{L}}{T_{H}}$ (a) $\eta=1-(80 / 100)=20 \%$ (b) $\eta=1-(20 / 40)=50 \%$ $\text { (c) } \eta=1-(60 / 80)=25 \%$ $\text { (d) } \eta=1-(40 / 60)=33 \%$ $\therefore$ The least efficiency of carnot's engine is in case of option (a).
Karnataka CET-2017
Thermodynamics
148587
A Carnot engine working between $300 \mathrm{~K}$ and $400 \mathrm{~K}$ has $800 \mathrm{~J}$ of useful work. The amount of heat energy supplied to the engine from the source is :
1 $2400 \mathrm{~J}$
2 $3200 \mathrm{~J}$
3 $1200 \mathrm{~J}$
4 $3600 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{H}}=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=300 \mathrm{~K}$ $\mathrm{~W}=800 \mathrm{~J}$ The efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}=\frac{W}{Q_{H}}$ $1-\frac{300}{400}=\frac{800}{Q_{H}}$ $\frac{1}{4}=\frac{800}{Q_{H}}$ $\Rightarrow \quad Q_{H}=3200 \mathrm{~J}$
Karnataka CET-2016
Thermodynamics
148588
The efficiency of a Carnot engine which operates between the two temperature $T_{1}=\mathbf{5 0 0}$ $K$ and $T_{2}=300 \mathrm{~K}$ is :
1 $75 \%$
2 $50 \%$
3 $40 \%$
4 $25 \%$
Explanation:
C Given that, $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ Efficiency of carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $\therefore \quad \eta =\left[1-\frac{300}{500}\right] \times 100$ $=0.4 \times 100$ $\eta =40 \%$
Karnataka CET-2015
Thermodynamics
148589
What is the source temperature of the Carnot engine required to get $70 \%$ efficiency? Given, sink temperature $=27^{\circ} \mathrm{C}$.
1 $1000^{\circ} \mathrm{C}$
2 $90^{\circ} \mathrm{C}$
3 $270^{\circ} \mathrm{C}$
4 $727^{\circ} \mathrm{C}$
Explanation:
D Given, Efficiency $\eta=70 \%$ Sink temperature $\left(T_{2}\right)=27^{\circ} \mathrm{C}$ $=27+273=300 \mathrm{~K}$ Source temperature $\left(\mathrm{T}_{1}\right)=$ ? As we know that efficiency is given by $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ \(70 \% =1-\frac{300}{\mathrm{~T}_1}\) \(\frac{70}{100} =1-\frac{300}{\mathrm{~T}_1}\) \(0.7 =1-\frac{300}{\mathrm{~T}_1}\) \(\frac{300}{\mathrm{~T}_1} =1-0.7\) \(\mathrm{~T}_1 =\frac{300}{0.3}\) \(\mathrm{~T}_1 =1000 \mathrm{~K}\) \(\mathrm{~T}_1 =1000-273=727^{\circ} \mathrm{C}\)
148586
For which combination of working temperatures, the efficiency of Carnot's engine is the least ?
1 $100 \mathrm{~K}, 80 \mathrm{~K}$
2 $40 \mathrm{~K}, 20 \mathrm{~K}$
3 $80 \mathrm{~K}, 60 \mathrm{~K}$
4 $60 \mathrm{~K}, 40 \mathrm{~K}$
Explanation:
A We know that the efficiency of Carnot's Engine $(\eta)=1-\frac{T_{L}}{T_{H}}$ (a) $\eta=1-(80 / 100)=20 \%$ (b) $\eta=1-(20 / 40)=50 \%$ $\text { (c) } \eta=1-(60 / 80)=25 \%$ $\text { (d) } \eta=1-(40 / 60)=33 \%$ $\therefore$ The least efficiency of carnot's engine is in case of option (a).
Karnataka CET-2017
Thermodynamics
148587
A Carnot engine working between $300 \mathrm{~K}$ and $400 \mathrm{~K}$ has $800 \mathrm{~J}$ of useful work. The amount of heat energy supplied to the engine from the source is :
1 $2400 \mathrm{~J}$
2 $3200 \mathrm{~J}$
3 $1200 \mathrm{~J}$
4 $3600 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{H}}=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=300 \mathrm{~K}$ $\mathrm{~W}=800 \mathrm{~J}$ The efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}=\frac{W}{Q_{H}}$ $1-\frac{300}{400}=\frac{800}{Q_{H}}$ $\frac{1}{4}=\frac{800}{Q_{H}}$ $\Rightarrow \quad Q_{H}=3200 \mathrm{~J}$
Karnataka CET-2016
Thermodynamics
148588
The efficiency of a Carnot engine which operates between the two temperature $T_{1}=\mathbf{5 0 0}$ $K$ and $T_{2}=300 \mathrm{~K}$ is :
1 $75 \%$
2 $50 \%$
3 $40 \%$
4 $25 \%$
Explanation:
C Given that, $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ Efficiency of carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $\therefore \quad \eta =\left[1-\frac{300}{500}\right] \times 100$ $=0.4 \times 100$ $\eta =40 \%$
Karnataka CET-2015
Thermodynamics
148589
What is the source temperature of the Carnot engine required to get $70 \%$ efficiency? Given, sink temperature $=27^{\circ} \mathrm{C}$.
1 $1000^{\circ} \mathrm{C}$
2 $90^{\circ} \mathrm{C}$
3 $270^{\circ} \mathrm{C}$
4 $727^{\circ} \mathrm{C}$
Explanation:
D Given, Efficiency $\eta=70 \%$ Sink temperature $\left(T_{2}\right)=27^{\circ} \mathrm{C}$ $=27+273=300 \mathrm{~K}$ Source temperature $\left(\mathrm{T}_{1}\right)=$ ? As we know that efficiency is given by $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ \(70 \% =1-\frac{300}{\mathrm{~T}_1}\) \(\frac{70}{100} =1-\frac{300}{\mathrm{~T}_1}\) \(0.7 =1-\frac{300}{\mathrm{~T}_1}\) \(\frac{300}{\mathrm{~T}_1} =1-0.7\) \(\mathrm{~T}_1 =\frac{300}{0.3}\) \(\mathrm{~T}_1 =1000 \mathrm{~K}\) \(\mathrm{~T}_1 =1000-273=727^{\circ} \mathrm{C}\)
148586
For which combination of working temperatures, the efficiency of Carnot's engine is the least ?
1 $100 \mathrm{~K}, 80 \mathrm{~K}$
2 $40 \mathrm{~K}, 20 \mathrm{~K}$
3 $80 \mathrm{~K}, 60 \mathrm{~K}$
4 $60 \mathrm{~K}, 40 \mathrm{~K}$
Explanation:
A We know that the efficiency of Carnot's Engine $(\eta)=1-\frac{T_{L}}{T_{H}}$ (a) $\eta=1-(80 / 100)=20 \%$ (b) $\eta=1-(20 / 40)=50 \%$ $\text { (c) } \eta=1-(60 / 80)=25 \%$ $\text { (d) } \eta=1-(40 / 60)=33 \%$ $\therefore$ The least efficiency of carnot's engine is in case of option (a).
Karnataka CET-2017
Thermodynamics
148587
A Carnot engine working between $300 \mathrm{~K}$ and $400 \mathrm{~K}$ has $800 \mathrm{~J}$ of useful work. The amount of heat energy supplied to the engine from the source is :
1 $2400 \mathrm{~J}$
2 $3200 \mathrm{~J}$
3 $1200 \mathrm{~J}$
4 $3600 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{H}}=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=300 \mathrm{~K}$ $\mathrm{~W}=800 \mathrm{~J}$ The efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}=\frac{W}{Q_{H}}$ $1-\frac{300}{400}=\frac{800}{Q_{H}}$ $\frac{1}{4}=\frac{800}{Q_{H}}$ $\Rightarrow \quad Q_{H}=3200 \mathrm{~J}$
Karnataka CET-2016
Thermodynamics
148588
The efficiency of a Carnot engine which operates between the two temperature $T_{1}=\mathbf{5 0 0}$ $K$ and $T_{2}=300 \mathrm{~K}$ is :
1 $75 \%$
2 $50 \%$
3 $40 \%$
4 $25 \%$
Explanation:
C Given that, $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ Efficiency of carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $\therefore \quad \eta =\left[1-\frac{300}{500}\right] \times 100$ $=0.4 \times 100$ $\eta =40 \%$
Karnataka CET-2015
Thermodynamics
148589
What is the source temperature of the Carnot engine required to get $70 \%$ efficiency? Given, sink temperature $=27^{\circ} \mathrm{C}$.
1 $1000^{\circ} \mathrm{C}$
2 $90^{\circ} \mathrm{C}$
3 $270^{\circ} \mathrm{C}$
4 $727^{\circ} \mathrm{C}$
Explanation:
D Given, Efficiency $\eta=70 \%$ Sink temperature $\left(T_{2}\right)=27^{\circ} \mathrm{C}$ $=27+273=300 \mathrm{~K}$ Source temperature $\left(\mathrm{T}_{1}\right)=$ ? As we know that efficiency is given by $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ \(70 \% =1-\frac{300}{\mathrm{~T}_1}\) \(\frac{70}{100} =1-\frac{300}{\mathrm{~T}_1}\) \(0.7 =1-\frac{300}{\mathrm{~T}_1}\) \(\frac{300}{\mathrm{~T}_1} =1-0.7\) \(\mathrm{~T}_1 =\frac{300}{0.3}\) \(\mathrm{~T}_1 =1000 \mathrm{~K}\) \(\mathrm{~T}_1 =1000-273=727^{\circ} \mathrm{C}\)
148586
For which combination of working temperatures, the efficiency of Carnot's engine is the least ?
1 $100 \mathrm{~K}, 80 \mathrm{~K}$
2 $40 \mathrm{~K}, 20 \mathrm{~K}$
3 $80 \mathrm{~K}, 60 \mathrm{~K}$
4 $60 \mathrm{~K}, 40 \mathrm{~K}$
Explanation:
A We know that the efficiency of Carnot's Engine $(\eta)=1-\frac{T_{L}}{T_{H}}$ (a) $\eta=1-(80 / 100)=20 \%$ (b) $\eta=1-(20 / 40)=50 \%$ $\text { (c) } \eta=1-(60 / 80)=25 \%$ $\text { (d) } \eta=1-(40 / 60)=33 \%$ $\therefore$ The least efficiency of carnot's engine is in case of option (a).
Karnataka CET-2017
Thermodynamics
148587
A Carnot engine working between $300 \mathrm{~K}$ and $400 \mathrm{~K}$ has $800 \mathrm{~J}$ of useful work. The amount of heat energy supplied to the engine from the source is :
1 $2400 \mathrm{~J}$
2 $3200 \mathrm{~J}$
3 $1200 \mathrm{~J}$
4 $3600 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{H}}=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=300 \mathrm{~K}$ $\mathrm{~W}=800 \mathrm{~J}$ The efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}=\frac{W}{Q_{H}}$ $1-\frac{300}{400}=\frac{800}{Q_{H}}$ $\frac{1}{4}=\frac{800}{Q_{H}}$ $\Rightarrow \quad Q_{H}=3200 \mathrm{~J}$
Karnataka CET-2016
Thermodynamics
148588
The efficiency of a Carnot engine which operates between the two temperature $T_{1}=\mathbf{5 0 0}$ $K$ and $T_{2}=300 \mathrm{~K}$ is :
1 $75 \%$
2 $50 \%$
3 $40 \%$
4 $25 \%$
Explanation:
C Given that, $\mathrm{T}_{1}=500 \mathrm{~K}, \mathrm{~T}_{2}=300 \mathrm{~K}$ Efficiency of carnot's engine $(\eta)=1-\frac{T_{2}}{T_{1}}$ $\therefore \quad \eta =\left[1-\frac{300}{500}\right] \times 100$ $=0.4 \times 100$ $\eta =40 \%$
Karnataka CET-2015
Thermodynamics
148589
What is the source temperature of the Carnot engine required to get $70 \%$ efficiency? Given, sink temperature $=27^{\circ} \mathrm{C}$.
1 $1000^{\circ} \mathrm{C}$
2 $90^{\circ} \mathrm{C}$
3 $270^{\circ} \mathrm{C}$
4 $727^{\circ} \mathrm{C}$
Explanation:
D Given, Efficiency $\eta=70 \%$ Sink temperature $\left(T_{2}\right)=27^{\circ} \mathrm{C}$ $=27+273=300 \mathrm{~K}$ Source temperature $\left(\mathrm{T}_{1}\right)=$ ? As we know that efficiency is given by $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ \(70 \% =1-\frac{300}{\mathrm{~T}_1}\) \(\frac{70}{100} =1-\frac{300}{\mathrm{~T}_1}\) \(0.7 =1-\frac{300}{\mathrm{~T}_1}\) \(\frac{300}{\mathrm{~T}_1} =1-0.7\) \(\mathrm{~T}_1 =\frac{300}{0.3}\) \(\mathrm{~T}_1 =1000 \mathrm{~K}\) \(\mathrm{~T}_1 =1000-273=727^{\circ} \mathrm{C}\)
