148596
The temperature of the sink of a Carnot engine is $27^{\circ} \mathrm{C}$ and its efficiency is $25 \%$. Then temperature of the source is
1 $227^{\circ} \mathrm{C}$
2 $27^{\circ} \mathrm{C}$
3 $327^{\circ} \mathrm{C}$
4 $127^{\circ} \mathrm{C}$
Explanation:
D Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\eta=25 \%=\frac{25}{100}=0.25$ $\mathrm{T}_{1}=\text { ? }$ Efficiency of a carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ Where $T_{1}$ is the temperature of the source and $T_{2}$ is the temperature of the sink. $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $0.25=1-\frac{300}{\mathrm{~T}_{1}}$ $0.75=\frac{300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{~T}_{1}=400-273$ $\mathrm{~T}_{1}=127^{\circ} \mathrm{C}$
J and K CET- 2011
Thermodynamics
148598
A Carnot's engine working between $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ has a work output of $200 \mathrm{~J}$ per cycle. The energy supplied to the engine from the source in each cycle is
1 $400 \mathrm{~J}$
2 $800 \mathrm{~J}$
3 $600 \mathrm{~J}$
4 $500 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{T}_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ $\text { Work done per cycle } \mathrm{W}=200 \mathrm{~J}$ The efficiency of engine is defined as the ratio of work done to the heat supplied $\eta=\frac{W}{Q}=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{400}=\frac{1}{4}$ Since $\eta=\frac{W}{Q}$ $\frac{1}{4}=\frac{200}{Q}$ $Q=800 \mathrm{~J}$
J and K CET- 2009
Thermodynamics
148599
An ideal gas heat engine operates in a Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^{4}$ cals at the higher temperature. The amount of heat converted into work is
1 $4.8 \times 10^{4} \mathrm{Cal}$
2 $1.2 \times 10^{4} \mathrm{Cal}$
3 $3.5 \times 10^{4} \mathrm{Cal}$
4 $1.6 \times 10^{4} \mathrm{Cal}$
Explanation:
B Given that, $\mathrm{T}_{2}=127^{\circ} \mathrm{C}=(127+273) \mathrm{K}=400 \mathrm{~K}$ $\mathrm{T}_{1}=227^{\circ} \mathrm{C}=(227+273) \mathrm{K}=500 \mathrm{~K}$ For Carnot cycle, $\frac{\mathrm{W}}{\mathrm{Q}_{\mathrm{h}}}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ The amount of heat converted into work $\mathrm{W}=\left(1-\frac{400}{500}\right) \times 6 \times 10^{4}$ $\mathrm{W}=1.2 \times 10^{4} \mathrm{Cal} \text {. }$
AP EAMCET-03.09.2021
Thermodynamics
148600
The temperatures $T_{1}$ and $T_{2}$ of heat reservoirs in the ideal Carnot engine are $1500^{\circ} \mathrm{C}$ and $500^{\circ} \mathrm{C}$ respectively. If $\mathrm{T}_{1}$ increases by $100^{\circ} \mathrm{C}$. What will be the efficiency of the engine?
1 $62 \%$
2 $59 \%$
3 $95 \%$
4 $100 \%$
Explanation:
B Given that, $\mathrm{T}_{1}=1500^{\circ} \mathrm{C}=1500+273=1773 \mathrm{~K}$ $\mathrm{~T}_{2}=500^{\circ} \mathrm{C}=500+273=773 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}=\frac{T_{1}-T_{2}}{T_{1}}$ It is clearly given that $\mathrm{T}_{1}$ is increased by $100^{\circ} \mathrm{C}$ $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ $\mathrm{T}_{1}=1500^{\circ} \mathrm{C}+100^{\circ} \mathrm{C}=1600^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=(1600+273) \mathrm{K}=1873 \mathrm{~K}$ $\eta=\frac{1873-773}{1873}=\frac{1100}{1873}=0.587 \simeq 0.59$ $\eta=59 \%$
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Thermodynamics
148596
The temperature of the sink of a Carnot engine is $27^{\circ} \mathrm{C}$ and its efficiency is $25 \%$. Then temperature of the source is
1 $227^{\circ} \mathrm{C}$
2 $27^{\circ} \mathrm{C}$
3 $327^{\circ} \mathrm{C}$
4 $127^{\circ} \mathrm{C}$
Explanation:
D Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\eta=25 \%=\frac{25}{100}=0.25$ $\mathrm{T}_{1}=\text { ? }$ Efficiency of a carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ Where $T_{1}$ is the temperature of the source and $T_{2}$ is the temperature of the sink. $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $0.25=1-\frac{300}{\mathrm{~T}_{1}}$ $0.75=\frac{300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{~T}_{1}=400-273$ $\mathrm{~T}_{1}=127^{\circ} \mathrm{C}$
J and K CET- 2011
Thermodynamics
148598
A Carnot's engine working between $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ has a work output of $200 \mathrm{~J}$ per cycle. The energy supplied to the engine from the source in each cycle is
1 $400 \mathrm{~J}$
2 $800 \mathrm{~J}$
3 $600 \mathrm{~J}$
4 $500 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{T}_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ $\text { Work done per cycle } \mathrm{W}=200 \mathrm{~J}$ The efficiency of engine is defined as the ratio of work done to the heat supplied $\eta=\frac{W}{Q}=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{400}=\frac{1}{4}$ Since $\eta=\frac{W}{Q}$ $\frac{1}{4}=\frac{200}{Q}$ $Q=800 \mathrm{~J}$
J and K CET- 2009
Thermodynamics
148599
An ideal gas heat engine operates in a Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^{4}$ cals at the higher temperature. The amount of heat converted into work is
1 $4.8 \times 10^{4} \mathrm{Cal}$
2 $1.2 \times 10^{4} \mathrm{Cal}$
3 $3.5 \times 10^{4} \mathrm{Cal}$
4 $1.6 \times 10^{4} \mathrm{Cal}$
Explanation:
B Given that, $\mathrm{T}_{2}=127^{\circ} \mathrm{C}=(127+273) \mathrm{K}=400 \mathrm{~K}$ $\mathrm{T}_{1}=227^{\circ} \mathrm{C}=(227+273) \mathrm{K}=500 \mathrm{~K}$ For Carnot cycle, $\frac{\mathrm{W}}{\mathrm{Q}_{\mathrm{h}}}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ The amount of heat converted into work $\mathrm{W}=\left(1-\frac{400}{500}\right) \times 6 \times 10^{4}$ $\mathrm{W}=1.2 \times 10^{4} \mathrm{Cal} \text {. }$
AP EAMCET-03.09.2021
Thermodynamics
148600
The temperatures $T_{1}$ and $T_{2}$ of heat reservoirs in the ideal Carnot engine are $1500^{\circ} \mathrm{C}$ and $500^{\circ} \mathrm{C}$ respectively. If $\mathrm{T}_{1}$ increases by $100^{\circ} \mathrm{C}$. What will be the efficiency of the engine?
1 $62 \%$
2 $59 \%$
3 $95 \%$
4 $100 \%$
Explanation:
B Given that, $\mathrm{T}_{1}=1500^{\circ} \mathrm{C}=1500+273=1773 \mathrm{~K}$ $\mathrm{~T}_{2}=500^{\circ} \mathrm{C}=500+273=773 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}=\frac{T_{1}-T_{2}}{T_{1}}$ It is clearly given that $\mathrm{T}_{1}$ is increased by $100^{\circ} \mathrm{C}$ $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ $\mathrm{T}_{1}=1500^{\circ} \mathrm{C}+100^{\circ} \mathrm{C}=1600^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=(1600+273) \mathrm{K}=1873 \mathrm{~K}$ $\eta=\frac{1873-773}{1873}=\frac{1100}{1873}=0.587 \simeq 0.59$ $\eta=59 \%$
148596
The temperature of the sink of a Carnot engine is $27^{\circ} \mathrm{C}$ and its efficiency is $25 \%$. Then temperature of the source is
1 $227^{\circ} \mathrm{C}$
2 $27^{\circ} \mathrm{C}$
3 $327^{\circ} \mathrm{C}$
4 $127^{\circ} \mathrm{C}$
Explanation:
D Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\eta=25 \%=\frac{25}{100}=0.25$ $\mathrm{T}_{1}=\text { ? }$ Efficiency of a carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ Where $T_{1}$ is the temperature of the source and $T_{2}$ is the temperature of the sink. $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $0.25=1-\frac{300}{\mathrm{~T}_{1}}$ $0.75=\frac{300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{~T}_{1}=400-273$ $\mathrm{~T}_{1}=127^{\circ} \mathrm{C}$
J and K CET- 2011
Thermodynamics
148598
A Carnot's engine working between $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ has a work output of $200 \mathrm{~J}$ per cycle. The energy supplied to the engine from the source in each cycle is
1 $400 \mathrm{~J}$
2 $800 \mathrm{~J}$
3 $600 \mathrm{~J}$
4 $500 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{T}_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ $\text { Work done per cycle } \mathrm{W}=200 \mathrm{~J}$ The efficiency of engine is defined as the ratio of work done to the heat supplied $\eta=\frac{W}{Q}=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{400}=\frac{1}{4}$ Since $\eta=\frac{W}{Q}$ $\frac{1}{4}=\frac{200}{Q}$ $Q=800 \mathrm{~J}$
J and K CET- 2009
Thermodynamics
148599
An ideal gas heat engine operates in a Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^{4}$ cals at the higher temperature. The amount of heat converted into work is
1 $4.8 \times 10^{4} \mathrm{Cal}$
2 $1.2 \times 10^{4} \mathrm{Cal}$
3 $3.5 \times 10^{4} \mathrm{Cal}$
4 $1.6 \times 10^{4} \mathrm{Cal}$
Explanation:
B Given that, $\mathrm{T}_{2}=127^{\circ} \mathrm{C}=(127+273) \mathrm{K}=400 \mathrm{~K}$ $\mathrm{T}_{1}=227^{\circ} \mathrm{C}=(227+273) \mathrm{K}=500 \mathrm{~K}$ For Carnot cycle, $\frac{\mathrm{W}}{\mathrm{Q}_{\mathrm{h}}}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ The amount of heat converted into work $\mathrm{W}=\left(1-\frac{400}{500}\right) \times 6 \times 10^{4}$ $\mathrm{W}=1.2 \times 10^{4} \mathrm{Cal} \text {. }$
AP EAMCET-03.09.2021
Thermodynamics
148600
The temperatures $T_{1}$ and $T_{2}$ of heat reservoirs in the ideal Carnot engine are $1500^{\circ} \mathrm{C}$ and $500^{\circ} \mathrm{C}$ respectively. If $\mathrm{T}_{1}$ increases by $100^{\circ} \mathrm{C}$. What will be the efficiency of the engine?
1 $62 \%$
2 $59 \%$
3 $95 \%$
4 $100 \%$
Explanation:
B Given that, $\mathrm{T}_{1}=1500^{\circ} \mathrm{C}=1500+273=1773 \mathrm{~K}$ $\mathrm{~T}_{2}=500^{\circ} \mathrm{C}=500+273=773 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}=\frac{T_{1}-T_{2}}{T_{1}}$ It is clearly given that $\mathrm{T}_{1}$ is increased by $100^{\circ} \mathrm{C}$ $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ $\mathrm{T}_{1}=1500^{\circ} \mathrm{C}+100^{\circ} \mathrm{C}=1600^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=(1600+273) \mathrm{K}=1873 \mathrm{~K}$ $\eta=\frac{1873-773}{1873}=\frac{1100}{1873}=0.587 \simeq 0.59$ $\eta=59 \%$
148596
The temperature of the sink of a Carnot engine is $27^{\circ} \mathrm{C}$ and its efficiency is $25 \%$. Then temperature of the source is
1 $227^{\circ} \mathrm{C}$
2 $27^{\circ} \mathrm{C}$
3 $327^{\circ} \mathrm{C}$
4 $127^{\circ} \mathrm{C}$
Explanation:
D Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\eta=25 \%=\frac{25}{100}=0.25$ $\mathrm{T}_{1}=\text { ? }$ Efficiency of a carnot engine $\eta=1-\frac{T_{2}}{T_{1}}$ Where $T_{1}$ is the temperature of the source and $T_{2}$ is the temperature of the sink. $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $0.25=1-\frac{300}{\mathrm{~T}_{1}}$ $0.75=\frac{300}{\mathrm{~T}_{1}}$ $\mathrm{~T}_{1}=\frac{300}{0.75}=400 \mathrm{~K}$ $\mathrm{~T}_{1}=400-273$ $\mathrm{~T}_{1}=127^{\circ} \mathrm{C}$
J and K CET- 2011
Thermodynamics
148598
A Carnot's engine working between $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ has a work output of $200 \mathrm{~J}$ per cycle. The energy supplied to the engine from the source in each cycle is
1 $400 \mathrm{~J}$
2 $800 \mathrm{~J}$
3 $600 \mathrm{~J}$
4 $500 \mathrm{~J}$
Explanation:
B Given that, $\mathrm{T}_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{T}_{1}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ $\text { Work done per cycle } \mathrm{W}=200 \mathrm{~J}$ The efficiency of engine is defined as the ratio of work done to the heat supplied $\eta=\frac{W}{Q}=1-\frac{T_{2}}{T_{1}}$ $\eta=1-\frac{300}{400}=\frac{1}{4}$ Since $\eta=\frac{W}{Q}$ $\frac{1}{4}=\frac{200}{Q}$ $Q=800 \mathrm{~J}$
J and K CET- 2009
Thermodynamics
148599
An ideal gas heat engine operates in a Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^{4}$ cals at the higher temperature. The amount of heat converted into work is
1 $4.8 \times 10^{4} \mathrm{Cal}$
2 $1.2 \times 10^{4} \mathrm{Cal}$
3 $3.5 \times 10^{4} \mathrm{Cal}$
4 $1.6 \times 10^{4} \mathrm{Cal}$
Explanation:
B Given that, $\mathrm{T}_{2}=127^{\circ} \mathrm{C}=(127+273) \mathrm{K}=400 \mathrm{~K}$ $\mathrm{T}_{1}=227^{\circ} \mathrm{C}=(227+273) \mathrm{K}=500 \mathrm{~K}$ For Carnot cycle, $\frac{\mathrm{W}}{\mathrm{Q}_{\mathrm{h}}}=1-\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}$ The amount of heat converted into work $\mathrm{W}=\left(1-\frac{400}{500}\right) \times 6 \times 10^{4}$ $\mathrm{W}=1.2 \times 10^{4} \mathrm{Cal} \text {. }$
AP EAMCET-03.09.2021
Thermodynamics
148600
The temperatures $T_{1}$ and $T_{2}$ of heat reservoirs in the ideal Carnot engine are $1500^{\circ} \mathrm{C}$ and $500^{\circ} \mathrm{C}$ respectively. If $\mathrm{T}_{1}$ increases by $100^{\circ} \mathrm{C}$. What will be the efficiency of the engine?
1 $62 \%$
2 $59 \%$
3 $95 \%$
4 $100 \%$
Explanation:
B Given that, $\mathrm{T}_{1}=1500^{\circ} \mathrm{C}=1500+273=1773 \mathrm{~K}$ $\mathrm{~T}_{2}=500^{\circ} \mathrm{C}=500+273=773 \mathrm{~K}$ We know, efficiency of carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}=\frac{T_{1}-T_{2}}{T_{1}}$ It is clearly given that $\mathrm{T}_{1}$ is increased by $100^{\circ} \mathrm{C}$ $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ $\mathrm{T}_{1}=1500^{\circ} \mathrm{C}+100^{\circ} \mathrm{C}=1600^{\circ} \mathrm{C}$ $\mathrm{T}_{1}=(1600+273) \mathrm{K}=1873 \mathrm{~K}$ $\eta=\frac{1873-773}{1873}=\frac{1100}{1873}=0.587 \simeq 0.59$ $\eta=59 \%$