371371
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum?
1 \(600\;K,\,400\;K\)
2 \(400\;K,\,200\;K\)
3 \(500\;K,300\;K\)
4 \(300\;K,100\;K\)
Explanation:
Efficiency of Carnot's engine, \(\Rightarrow \eta=1-\dfrac{T_{2}}{T_{1}}\) \(\eta {\rm{ }}\) is maximum for \(300k,100k\)
KCET - 2013
PHXI12:THERMODYNAMICS
371372
A Carnot's engine operates with source at \(127^\circ C\) and sink at \(27^\circ C\). If the source supplies \(40\;kJ\) of heat energy, the work done by the engine is
371373
The efficiency of a carnot engine working between source temperature \(T\) and sink temperature \(27^\circ C\) is \(25{\rm{ }}\,\% \). The source temperature \(T\) is :
371374
Carnot engine cannot give \(100 \%\) efficiency, because we cannot
1 eliminate friction
2 find ideal sources
3 prevent radiation
4 reach absolute zero temperature
Explanation:
The efficiency of Carnot engine is \(\eta=\dfrac{T_{1}-T_{2}}{T_{1}}=1-\dfrac{T_{2}}{T_{1}}\left(T_{1}>T_{2}\right)\) where, \(T_{1}\) = source temperature and \(T_{2}=\text { sink temperature. }\) The value of efficiency of Carnot engine can be 1 or \(100 \%\), if the sink temperature \(\left(T_{2}\right)\) is at \(0 K\). In order to achieve \(100 \%\), efficiency \((\eta=1), Q_{2}\) must be equal to 0 which means all the heat from the source is converted to work. As practically, absolute zero temperature cannot be achieved. Hence, Carnot engine cannot give \(100 \%\) efficiency.
PHXI12:THERMODYNAMICS
371375
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 temperature of source and sink of the second engine are respectively
1 \(2\;{T_1},\,2\;{T_2}\)
2 \(2\;{T_1},\,\frac{{{T_2}}}{2}\)
3 \({T_1} + 5,\;\,{T_2} - 5\)
4 \({T_1} + 10,{\rm{ }}{T_2} - 10\)
Explanation:
Efficiency of Carnot's engine is \(\eta=1-\dfrac{T_{2}}{T_{1}}\) For first engine, \(0.5=1-\dfrac{T_{2}}{T_{1}}\) For second engine, \(0.5=1-\dfrac{T_{2}^{1}}{T_{1}^{1}}\) Efficiency remains the same when both \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are increased by same factor. Therefore, the temperature of source and sink of second engine are \(T_{1}^{\prime}=2 T_{1}, T_{2}^{\prime}=2 T_{2}\)
371371
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum?
1 \(600\;K,\,400\;K\)
2 \(400\;K,\,200\;K\)
3 \(500\;K,300\;K\)
4 \(300\;K,100\;K\)
Explanation:
Efficiency of Carnot's engine, \(\Rightarrow \eta=1-\dfrac{T_{2}}{T_{1}}\) \(\eta {\rm{ }}\) is maximum for \(300k,100k\)
KCET - 2013
PHXI12:THERMODYNAMICS
371372
A Carnot's engine operates with source at \(127^\circ C\) and sink at \(27^\circ C\). If the source supplies \(40\;kJ\) of heat energy, the work done by the engine is
371373
The efficiency of a carnot engine working between source temperature \(T\) and sink temperature \(27^\circ C\) is \(25{\rm{ }}\,\% \). The source temperature \(T\) is :
371374
Carnot engine cannot give \(100 \%\) efficiency, because we cannot
1 eliminate friction
2 find ideal sources
3 prevent radiation
4 reach absolute zero temperature
Explanation:
The efficiency of Carnot engine is \(\eta=\dfrac{T_{1}-T_{2}}{T_{1}}=1-\dfrac{T_{2}}{T_{1}}\left(T_{1}>T_{2}\right)\) where, \(T_{1}\) = source temperature and \(T_{2}=\text { sink temperature. }\) The value of efficiency of Carnot engine can be 1 or \(100 \%\), if the sink temperature \(\left(T_{2}\right)\) is at \(0 K\). In order to achieve \(100 \%\), efficiency \((\eta=1), Q_{2}\) must be equal to 0 which means all the heat from the source is converted to work. As practically, absolute zero temperature cannot be achieved. Hence, Carnot engine cannot give \(100 \%\) efficiency.
PHXI12:THERMODYNAMICS
371375
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 temperature of source and sink of the second engine are respectively
1 \(2\;{T_1},\,2\;{T_2}\)
2 \(2\;{T_1},\,\frac{{{T_2}}}{2}\)
3 \({T_1} + 5,\;\,{T_2} - 5\)
4 \({T_1} + 10,{\rm{ }}{T_2} - 10\)
Explanation:
Efficiency of Carnot's engine is \(\eta=1-\dfrac{T_{2}}{T_{1}}\) For first engine, \(0.5=1-\dfrac{T_{2}}{T_{1}}\) For second engine, \(0.5=1-\dfrac{T_{2}^{1}}{T_{1}^{1}}\) Efficiency remains the same when both \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are increased by same factor. Therefore, the temperature of source and sink of second engine are \(T_{1}^{\prime}=2 T_{1}, T_{2}^{\prime}=2 T_{2}\)
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PHXI12:THERMODYNAMICS
371371
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum?
1 \(600\;K,\,400\;K\)
2 \(400\;K,\,200\;K\)
3 \(500\;K,300\;K\)
4 \(300\;K,100\;K\)
Explanation:
Efficiency of Carnot's engine, \(\Rightarrow \eta=1-\dfrac{T_{2}}{T_{1}}\) \(\eta {\rm{ }}\) is maximum for \(300k,100k\)
KCET - 2013
PHXI12:THERMODYNAMICS
371372
A Carnot's engine operates with source at \(127^\circ C\) and sink at \(27^\circ C\). If the source supplies \(40\;kJ\) of heat energy, the work done by the engine is
371373
The efficiency of a carnot engine working between source temperature \(T\) and sink temperature \(27^\circ C\) is \(25{\rm{ }}\,\% \). The source temperature \(T\) is :
371374
Carnot engine cannot give \(100 \%\) efficiency, because we cannot
1 eliminate friction
2 find ideal sources
3 prevent radiation
4 reach absolute zero temperature
Explanation:
The efficiency of Carnot engine is \(\eta=\dfrac{T_{1}-T_{2}}{T_{1}}=1-\dfrac{T_{2}}{T_{1}}\left(T_{1}>T_{2}\right)\) where, \(T_{1}\) = source temperature and \(T_{2}=\text { sink temperature. }\) The value of efficiency of Carnot engine can be 1 or \(100 \%\), if the sink temperature \(\left(T_{2}\right)\) is at \(0 K\). In order to achieve \(100 \%\), efficiency \((\eta=1), Q_{2}\) must be equal to 0 which means all the heat from the source is converted to work. As practically, absolute zero temperature cannot be achieved. Hence, Carnot engine cannot give \(100 \%\) efficiency.
PHXI12:THERMODYNAMICS
371375
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 temperature of source and sink of the second engine are respectively
1 \(2\;{T_1},\,2\;{T_2}\)
2 \(2\;{T_1},\,\frac{{{T_2}}}{2}\)
3 \({T_1} + 5,\;\,{T_2} - 5\)
4 \({T_1} + 10,{\rm{ }}{T_2} - 10\)
Explanation:
Efficiency of Carnot's engine is \(\eta=1-\dfrac{T_{2}}{T_{1}}\) For first engine, \(0.5=1-\dfrac{T_{2}}{T_{1}}\) For second engine, \(0.5=1-\dfrac{T_{2}^{1}}{T_{1}^{1}}\) Efficiency remains the same when both \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are increased by same factor. Therefore, the temperature of source and sink of second engine are \(T_{1}^{\prime}=2 T_{1}, T_{2}^{\prime}=2 T_{2}\)
371371
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum?
1 \(600\;K,\,400\;K\)
2 \(400\;K,\,200\;K\)
3 \(500\;K,300\;K\)
4 \(300\;K,100\;K\)
Explanation:
Efficiency of Carnot's engine, \(\Rightarrow \eta=1-\dfrac{T_{2}}{T_{1}}\) \(\eta {\rm{ }}\) is maximum for \(300k,100k\)
KCET - 2013
PHXI12:THERMODYNAMICS
371372
A Carnot's engine operates with source at \(127^\circ C\) and sink at \(27^\circ C\). If the source supplies \(40\;kJ\) of heat energy, the work done by the engine is
371373
The efficiency of a carnot engine working between source temperature \(T\) and sink temperature \(27^\circ C\) is \(25{\rm{ }}\,\% \). The source temperature \(T\) is :
371374
Carnot engine cannot give \(100 \%\) efficiency, because we cannot
1 eliminate friction
2 find ideal sources
3 prevent radiation
4 reach absolute zero temperature
Explanation:
The efficiency of Carnot engine is \(\eta=\dfrac{T_{1}-T_{2}}{T_{1}}=1-\dfrac{T_{2}}{T_{1}}\left(T_{1}>T_{2}\right)\) where, \(T_{1}\) = source temperature and \(T_{2}=\text { sink temperature. }\) The value of efficiency of Carnot engine can be 1 or \(100 \%\), if the sink temperature \(\left(T_{2}\right)\) is at \(0 K\). In order to achieve \(100 \%\), efficiency \((\eta=1), Q_{2}\) must be equal to 0 which means all the heat from the source is converted to work. As practically, absolute zero temperature cannot be achieved. Hence, Carnot engine cannot give \(100 \%\) efficiency.
PHXI12:THERMODYNAMICS
371375
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 temperature of source and sink of the second engine are respectively
1 \(2\;{T_1},\,2\;{T_2}\)
2 \(2\;{T_1},\,\frac{{{T_2}}}{2}\)
3 \({T_1} + 5,\;\,{T_2} - 5\)
4 \({T_1} + 10,{\rm{ }}{T_2} - 10\)
Explanation:
Efficiency of Carnot's engine is \(\eta=1-\dfrac{T_{2}}{T_{1}}\) For first engine, \(0.5=1-\dfrac{T_{2}}{T_{1}}\) For second engine, \(0.5=1-\dfrac{T_{2}^{1}}{T_{1}^{1}}\) Efficiency remains the same when both \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are increased by same factor. Therefore, the temperature of source and sink of second engine are \(T_{1}^{\prime}=2 T_{1}, T_{2}^{\prime}=2 T_{2}\)
371371
For which combination of working temperatures of source and sink, the efficiency of Carnot's heat engine is maximum?
1 \(600\;K,\,400\;K\)
2 \(400\;K,\,200\;K\)
3 \(500\;K,300\;K\)
4 \(300\;K,100\;K\)
Explanation:
Efficiency of Carnot's engine, \(\Rightarrow \eta=1-\dfrac{T_{2}}{T_{1}}\) \(\eta {\rm{ }}\) is maximum for \(300k,100k\)
KCET - 2013
PHXI12:THERMODYNAMICS
371372
A Carnot's engine operates with source at \(127^\circ C\) and sink at \(27^\circ C\). If the source supplies \(40\;kJ\) of heat energy, the work done by the engine is
371373
The efficiency of a carnot engine working between source temperature \(T\) and sink temperature \(27^\circ C\) is \(25{\rm{ }}\,\% \). The source temperature \(T\) is :
371374
Carnot engine cannot give \(100 \%\) efficiency, because we cannot
1 eliminate friction
2 find ideal sources
3 prevent radiation
4 reach absolute zero temperature
Explanation:
The efficiency of Carnot engine is \(\eta=\dfrac{T_{1}-T_{2}}{T_{1}}=1-\dfrac{T_{2}}{T_{1}}\left(T_{1}>T_{2}\right)\) where, \(T_{1}\) = source temperature and \(T_{2}=\text { sink temperature. }\) The value of efficiency of Carnot engine can be 1 or \(100 \%\), if the sink temperature \(\left(T_{2}\right)\) is at \(0 K\). In order to achieve \(100 \%\), efficiency \((\eta=1), Q_{2}\) must be equal to 0 which means all the heat from the source is converted to work. As practically, absolute zero temperature cannot be achieved. Hence, Carnot engine cannot give \(100 \%\) efficiency.
PHXI12:THERMODYNAMICS
371375
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 temperature of source and sink of the second engine are respectively
1 \(2\;{T_1},\,2\;{T_2}\)
2 \(2\;{T_1},\,\frac{{{T_2}}}{2}\)
3 \({T_1} + 5,\;\,{T_2} - 5\)
4 \({T_1} + 10,{\rm{ }}{T_2} - 10\)
Explanation:
Efficiency of Carnot's engine is \(\eta=1-\dfrac{T_{2}}{T_{1}}\) For first engine, \(0.5=1-\dfrac{T_{2}}{T_{1}}\) For second engine, \(0.5=1-\dfrac{T_{2}^{1}}{T_{1}^{1}}\) Efficiency remains the same when both \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are increased by same factor. Therefore, the temperature of source and sink of second engine are \(T_{1}^{\prime}=2 T_{1}, T_{2}^{\prime}=2 T_{2}\)