148576
If sink and source temperature of a refrigerator are $4^{\circ} \mathrm{C}$ and $15^{\circ} \mathrm{C}$ respectively. Then efficiency of refrigerator is
1 0.076
2 0.0382
3 0.019
4 1
Explanation:
B Given, sink temperature of refrigerator $\left(\mathrm{T}_{2}\right)=$ $4^{\circ} \mathrm{C}$, source temperature of refrigerator $\left(\mathrm{T}_{1}\right)=15^{\circ} \mathrm{C}$ Efficiency of refrigerator is $\eta =\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)=\left(1-\frac{273+4}{273+15}\right)$ $=\left(1-\frac{277}{288}\right)=\frac{11}{288}$ $=0.0382$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148583
For a heat engine operating between temperature $t_{1}{ }^{\circ} \mathrm{C}$ and $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$, its efficiency will be
1 $\frac{t_{1}-t_{2}}{t_{2}}$
2 $\frac{t_{1}-t_{2}}{t_{1}+273}$
3 $\frac{t_{1}}{t_{2}}$
4 $1-\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}$
Explanation:
B The efficiency of the cycle $(\eta)=1-\frac{T_{2}}{T_{1}}$ Let, $\quad \mathrm{T}_{1}=\mathrm{t}_{1}+273$ $\mathrm{T}_{2}=\mathrm{t}_{2}+273$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}+273}$ $\eta=\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{\mathrm{t}_{1}+273}$
MHT-CET 2020
Thermodynamics
148591
The efficiency of Carnot's heat engine is 0.5 when the temperature of the source is $T_{1}$ and that of sink is $T_{2}$. The efficiency of another Carnot's heat engine is also 0.5 . the temperatures of source and sink of the second engine are respectively.
1 $2 \mathrm{~T}_{1}, 2 \mathrm{~T}_{2}$
2 $2 \mathrm{~T}_{1}, \frac{\mathrm{T}_{2}}{2}$
3 $\mathrm{T}_{1}+5, \mathrm{~T}_{2}-5$
4 $\mathrm{T}_{1}+10, \mathrm{~T}_{2}-10$
Explanation:
A We Know that efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $\mathrm{T}_{\mathrm{L}} \rightarrow$ low temperature (sink) $\mathrm{T}_{\mathrm{H}} \rightarrow$ high temperature (Source) Efficiency remains same when $T_{L}$ and $T_{H}$ are increased by same ratio.
Karnataka CET-2010
Thermodynamics
148601
The efficiency of a frictionless engine can be $100 \%$ if the temperature of the sink is
1 $0^{\circ} \mathrm{C}$
2 $0 \mathrm{~K}$
3 equal to that of source
4 less than that of the source
Explanation:
B The efficiency of a friction less engine can be $100 \%$. If the temperature of the sink is 0 Kelvin. Efficiency of engine $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $T_{L}$ is temperature of the sink. At $\mathrm{T}_{\mathrm{L}}=0 \mathrm{~K}$ (For friction less engine) $\eta=100 \%$
148576
If sink and source temperature of a refrigerator are $4^{\circ} \mathrm{C}$ and $15^{\circ} \mathrm{C}$ respectively. Then efficiency of refrigerator is
1 0.076
2 0.0382
3 0.019
4 1
Explanation:
B Given, sink temperature of refrigerator $\left(\mathrm{T}_{2}\right)=$ $4^{\circ} \mathrm{C}$, source temperature of refrigerator $\left(\mathrm{T}_{1}\right)=15^{\circ} \mathrm{C}$ Efficiency of refrigerator is $\eta =\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)=\left(1-\frac{273+4}{273+15}\right)$ $=\left(1-\frac{277}{288}\right)=\frac{11}{288}$ $=0.0382$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148583
For a heat engine operating between temperature $t_{1}{ }^{\circ} \mathrm{C}$ and $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$, its efficiency will be
1 $\frac{t_{1}-t_{2}}{t_{2}}$
2 $\frac{t_{1}-t_{2}}{t_{1}+273}$
3 $\frac{t_{1}}{t_{2}}$
4 $1-\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}$
Explanation:
B The efficiency of the cycle $(\eta)=1-\frac{T_{2}}{T_{1}}$ Let, $\quad \mathrm{T}_{1}=\mathrm{t}_{1}+273$ $\mathrm{T}_{2}=\mathrm{t}_{2}+273$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}+273}$ $\eta=\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{\mathrm{t}_{1}+273}$
MHT-CET 2020
Thermodynamics
148591
The efficiency of Carnot's heat engine is 0.5 when the temperature of the source is $T_{1}$ and that of sink is $T_{2}$. The efficiency of another Carnot's heat engine is also 0.5 . the temperatures of source and sink of the second engine are respectively.
1 $2 \mathrm{~T}_{1}, 2 \mathrm{~T}_{2}$
2 $2 \mathrm{~T}_{1}, \frac{\mathrm{T}_{2}}{2}$
3 $\mathrm{T}_{1}+5, \mathrm{~T}_{2}-5$
4 $\mathrm{T}_{1}+10, \mathrm{~T}_{2}-10$
Explanation:
A We Know that efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $\mathrm{T}_{\mathrm{L}} \rightarrow$ low temperature (sink) $\mathrm{T}_{\mathrm{H}} \rightarrow$ high temperature (Source) Efficiency remains same when $T_{L}$ and $T_{H}$ are increased by same ratio.
Karnataka CET-2010
Thermodynamics
148601
The efficiency of a frictionless engine can be $100 \%$ if the temperature of the sink is
1 $0^{\circ} \mathrm{C}$
2 $0 \mathrm{~K}$
3 equal to that of source
4 less than that of the source
Explanation:
B The efficiency of a friction less engine can be $100 \%$. If the temperature of the sink is 0 Kelvin. Efficiency of engine $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $T_{L}$ is temperature of the sink. At $\mathrm{T}_{\mathrm{L}}=0 \mathrm{~K}$ (For friction less engine) $\eta=100 \%$
148576
If sink and source temperature of a refrigerator are $4^{\circ} \mathrm{C}$ and $15^{\circ} \mathrm{C}$ respectively. Then efficiency of refrigerator is
1 0.076
2 0.0382
3 0.019
4 1
Explanation:
B Given, sink temperature of refrigerator $\left(\mathrm{T}_{2}\right)=$ $4^{\circ} \mathrm{C}$, source temperature of refrigerator $\left(\mathrm{T}_{1}\right)=15^{\circ} \mathrm{C}$ Efficiency of refrigerator is $\eta =\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)=\left(1-\frac{273+4}{273+15}\right)$ $=\left(1-\frac{277}{288}\right)=\frac{11}{288}$ $=0.0382$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148583
For a heat engine operating between temperature $t_{1}{ }^{\circ} \mathrm{C}$ and $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$, its efficiency will be
1 $\frac{t_{1}-t_{2}}{t_{2}}$
2 $\frac{t_{1}-t_{2}}{t_{1}+273}$
3 $\frac{t_{1}}{t_{2}}$
4 $1-\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}$
Explanation:
B The efficiency of the cycle $(\eta)=1-\frac{T_{2}}{T_{1}}$ Let, $\quad \mathrm{T}_{1}=\mathrm{t}_{1}+273$ $\mathrm{T}_{2}=\mathrm{t}_{2}+273$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}+273}$ $\eta=\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{\mathrm{t}_{1}+273}$
MHT-CET 2020
Thermodynamics
148591
The efficiency of Carnot's heat engine is 0.5 when the temperature of the source is $T_{1}$ and that of sink is $T_{2}$. The efficiency of another Carnot's heat engine is also 0.5 . the temperatures of source and sink of the second engine are respectively.
1 $2 \mathrm{~T}_{1}, 2 \mathrm{~T}_{2}$
2 $2 \mathrm{~T}_{1}, \frac{\mathrm{T}_{2}}{2}$
3 $\mathrm{T}_{1}+5, \mathrm{~T}_{2}-5$
4 $\mathrm{T}_{1}+10, \mathrm{~T}_{2}-10$
Explanation:
A We Know that efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $\mathrm{T}_{\mathrm{L}} \rightarrow$ low temperature (sink) $\mathrm{T}_{\mathrm{H}} \rightarrow$ high temperature (Source) Efficiency remains same when $T_{L}$ and $T_{H}$ are increased by same ratio.
Karnataka CET-2010
Thermodynamics
148601
The efficiency of a frictionless engine can be $100 \%$ if the temperature of the sink is
1 $0^{\circ} \mathrm{C}$
2 $0 \mathrm{~K}$
3 equal to that of source
4 less than that of the source
Explanation:
B The efficiency of a friction less engine can be $100 \%$. If the temperature of the sink is 0 Kelvin. Efficiency of engine $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $T_{L}$ is temperature of the sink. At $\mathrm{T}_{\mathrm{L}}=0 \mathrm{~K}$ (For friction less engine) $\eta=100 \%$
148576
If sink and source temperature of a refrigerator are $4^{\circ} \mathrm{C}$ and $15^{\circ} \mathrm{C}$ respectively. Then efficiency of refrigerator is
1 0.076
2 0.0382
3 0.019
4 1
Explanation:
B Given, sink temperature of refrigerator $\left(\mathrm{T}_{2}\right)=$ $4^{\circ} \mathrm{C}$, source temperature of refrigerator $\left(\mathrm{T}_{1}\right)=15^{\circ} \mathrm{C}$ Efficiency of refrigerator is $\eta =\left(1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)=\left(1-\frac{273+4}{273+15}\right)$ $=\left(1-\frac{277}{288}\right)=\frac{11}{288}$ $=0.0382$
AIIMS-25.05.2019(M) Shift-1
Thermodynamics
148583
For a heat engine operating between temperature $t_{1}{ }^{\circ} \mathrm{C}$ and $\mathrm{t}_{2}{ }^{\circ} \mathrm{C}$, its efficiency will be
1 $\frac{t_{1}-t_{2}}{t_{2}}$
2 $\frac{t_{1}-t_{2}}{t_{1}+273}$
3 $\frac{t_{1}}{t_{2}}$
4 $1-\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}$
Explanation:
B The efficiency of the cycle $(\eta)=1-\frac{T_{2}}{T_{1}}$ Let, $\quad \mathrm{T}_{1}=\mathrm{t}_{1}+273$ $\mathrm{T}_{2}=\mathrm{t}_{2}+273$ $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=1-\frac{\mathrm{t}_{2}+273}{\mathrm{t}_{1}+273}$ $\eta=\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{\mathrm{t}_{1}+273}$
MHT-CET 2020
Thermodynamics
148591
The efficiency of Carnot's heat engine is 0.5 when the temperature of the source is $T_{1}$ and that of sink is $T_{2}$. The efficiency of another Carnot's heat engine is also 0.5 . the temperatures of source and sink of the second engine are respectively.
1 $2 \mathrm{~T}_{1}, 2 \mathrm{~T}_{2}$
2 $2 \mathrm{~T}_{1}, \frac{\mathrm{T}_{2}}{2}$
3 $\mathrm{T}_{1}+5, \mathrm{~T}_{2}-5$
4 $\mathrm{T}_{1}+10, \mathrm{~T}_{2}-10$
Explanation:
A We Know that efficiency of carnot engine $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $\mathrm{T}_{\mathrm{L}} \rightarrow$ low temperature (sink) $\mathrm{T}_{\mathrm{H}} \rightarrow$ high temperature (Source) Efficiency remains same when $T_{L}$ and $T_{H}$ are increased by same ratio.
Karnataka CET-2010
Thermodynamics
148601
The efficiency of a frictionless engine can be $100 \%$ if the temperature of the sink is
1 $0^{\circ} \mathrm{C}$
2 $0 \mathrm{~K}$
3 equal to that of source
4 less than that of the source
Explanation:
B The efficiency of a friction less engine can be $100 \%$. If the temperature of the sink is 0 Kelvin. Efficiency of engine $\eta=1-\frac{T_{L}}{T_{H}}$ Where, $T_{L}$ is temperature of the sink. At $\mathrm{T}_{\mathrm{L}}=0 \mathrm{~K}$ (For friction less engine) $\eta=100 \%$
