371324
A reversible engine takes heat from a reservior at \(527^\circ C\) and gives out to the sink at \(127^\circ C\). The engine is required to perform useful mechanical work at the rate of \(750\;W\). How many calories per second must the engine take from the reservoir.
371326
An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are \({Q_1} = 5960\;J,\,{Q_2} = - 5585\;J,\) \({Q_3} = - 2980\;J\) and \({Q_4} = 3645\;J\) respectively. The efficiency of the cycle is
1 \(10.82\,\% \)
2 \(100\,\% \)
3 \(15{\rm{ }}\,\% \)
4 \(28\,\% \)
Explanation:
From the given problem, \(\Delta Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\) \(=5960-5585-2980+3645\) \(\Delta Q = 9605 - 8565 = 1040\;J\) Efficiency of a cycle is defined as \(\eta=\dfrac{\text { Network }}{\text { Input heat }}=\dfrac{\Delta W}{Q_{1}+Q_{4}}=\dfrac{\Delta Q}{Q_{1}+Q_{4}}\) Putting \(\Delta Q = 1040\;J\) and \({Q_2} + {Q_1} = 5960 + 3645 = 9605\;J\) \(\therefore \quad \eta = \frac{{1040}}{{9605}} = 0.1082 = 10.82\,\% \)
PHXI12:THERMODYNAMICS
371327
An ideal heat engine exhausting heat at \({77^{\circ} {C}}\) is to have \({30 \%}\) efficiency. It must take heat at
1 \({127^{\circ} {C}}\)
2 \({227^{\circ} {C}}\)
3 \({327^{\circ} {C}}\)
4 \({673^{\circ} {C}}\)
Explanation:
Given: \({\eta=30 \%=0.3}\) \({T_{2}=77^{\circ} {C}=350 {~K}, T_{1}=}\) ? Then \({\eta=1-\dfrac{T_{2}}{T_{1}} \quad \therefore \quad 0.3=1-\dfrac{350}{T_{1}}}\) \({\Rightarrow T_{1}=500 {~K}}\) or \({T_{1}=227^{\circ} {C}}\). So correct option is (2)
371324
A reversible engine takes heat from a reservior at \(527^\circ C\) and gives out to the sink at \(127^\circ C\). The engine is required to perform useful mechanical work at the rate of \(750\;W\). How many calories per second must the engine take from the reservoir.
371326
An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are \({Q_1} = 5960\;J,\,{Q_2} = - 5585\;J,\) \({Q_3} = - 2980\;J\) and \({Q_4} = 3645\;J\) respectively. The efficiency of the cycle is
1 \(10.82\,\% \)
2 \(100\,\% \)
3 \(15{\rm{ }}\,\% \)
4 \(28\,\% \)
Explanation:
From the given problem, \(\Delta Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\) \(=5960-5585-2980+3645\) \(\Delta Q = 9605 - 8565 = 1040\;J\) Efficiency of a cycle is defined as \(\eta=\dfrac{\text { Network }}{\text { Input heat }}=\dfrac{\Delta W}{Q_{1}+Q_{4}}=\dfrac{\Delta Q}{Q_{1}+Q_{4}}\) Putting \(\Delta Q = 1040\;J\) and \({Q_2} + {Q_1} = 5960 + 3645 = 9605\;J\) \(\therefore \quad \eta = \frac{{1040}}{{9605}} = 0.1082 = 10.82\,\% \)
PHXI12:THERMODYNAMICS
371327
An ideal heat engine exhausting heat at \({77^{\circ} {C}}\) is to have \({30 \%}\) efficiency. It must take heat at
1 \({127^{\circ} {C}}\)
2 \({227^{\circ} {C}}\)
3 \({327^{\circ} {C}}\)
4 \({673^{\circ} {C}}\)
Explanation:
Given: \({\eta=30 \%=0.3}\) \({T_{2}=77^{\circ} {C}=350 {~K}, T_{1}=}\) ? Then \({\eta=1-\dfrac{T_{2}}{T_{1}} \quad \therefore \quad 0.3=1-\dfrac{350}{T_{1}}}\) \({\Rightarrow T_{1}=500 {~K}}\) or \({T_{1}=227^{\circ} {C}}\). So correct option is (2)
371324
A reversible engine takes heat from a reservior at \(527^\circ C\) and gives out to the sink at \(127^\circ C\). The engine is required to perform useful mechanical work at the rate of \(750\;W\). How many calories per second must the engine take from the reservoir.
371326
An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are \({Q_1} = 5960\;J,\,{Q_2} = - 5585\;J,\) \({Q_3} = - 2980\;J\) and \({Q_4} = 3645\;J\) respectively. The efficiency of the cycle is
1 \(10.82\,\% \)
2 \(100\,\% \)
3 \(15{\rm{ }}\,\% \)
4 \(28\,\% \)
Explanation:
From the given problem, \(\Delta Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\) \(=5960-5585-2980+3645\) \(\Delta Q = 9605 - 8565 = 1040\;J\) Efficiency of a cycle is defined as \(\eta=\dfrac{\text { Network }}{\text { Input heat }}=\dfrac{\Delta W}{Q_{1}+Q_{4}}=\dfrac{\Delta Q}{Q_{1}+Q_{4}}\) Putting \(\Delta Q = 1040\;J\) and \({Q_2} + {Q_1} = 5960 + 3645 = 9605\;J\) \(\therefore \quad \eta = \frac{{1040}}{{9605}} = 0.1082 = 10.82\,\% \)
PHXI12:THERMODYNAMICS
371327
An ideal heat engine exhausting heat at \({77^{\circ} {C}}\) is to have \({30 \%}\) efficiency. It must take heat at
1 \({127^{\circ} {C}}\)
2 \({227^{\circ} {C}}\)
3 \({327^{\circ} {C}}\)
4 \({673^{\circ} {C}}\)
Explanation:
Given: \({\eta=30 \%=0.3}\) \({T_{2}=77^{\circ} {C}=350 {~K}, T_{1}=}\) ? Then \({\eta=1-\dfrac{T_{2}}{T_{1}} \quad \therefore \quad 0.3=1-\dfrac{350}{T_{1}}}\) \({\Rightarrow T_{1}=500 {~K}}\) or \({T_{1}=227^{\circ} {C}}\). So correct option is (2)
371324
A reversible engine takes heat from a reservior at \(527^\circ C\) and gives out to the sink at \(127^\circ C\). The engine is required to perform useful mechanical work at the rate of \(750\;W\). How many calories per second must the engine take from the reservoir.
371326
An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are \({Q_1} = 5960\;J,\,{Q_2} = - 5585\;J,\) \({Q_3} = - 2980\;J\) and \({Q_4} = 3645\;J\) respectively. The efficiency of the cycle is
1 \(10.82\,\% \)
2 \(100\,\% \)
3 \(15{\rm{ }}\,\% \)
4 \(28\,\% \)
Explanation:
From the given problem, \(\Delta Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\) \(=5960-5585-2980+3645\) \(\Delta Q = 9605 - 8565 = 1040\;J\) Efficiency of a cycle is defined as \(\eta=\dfrac{\text { Network }}{\text { Input heat }}=\dfrac{\Delta W}{Q_{1}+Q_{4}}=\dfrac{\Delta Q}{Q_{1}+Q_{4}}\) Putting \(\Delta Q = 1040\;J\) and \({Q_2} + {Q_1} = 5960 + 3645 = 9605\;J\) \(\therefore \quad \eta = \frac{{1040}}{{9605}} = 0.1082 = 10.82\,\% \)
PHXI12:THERMODYNAMICS
371327
An ideal heat engine exhausting heat at \({77^{\circ} {C}}\) is to have \({30 \%}\) efficiency. It must take heat at
1 \({127^{\circ} {C}}\)
2 \({227^{\circ} {C}}\)
3 \({327^{\circ} {C}}\)
4 \({673^{\circ} {C}}\)
Explanation:
Given: \({\eta=30 \%=0.3}\) \({T_{2}=77^{\circ} {C}=350 {~K}, T_{1}=}\) ? Then \({\eta=1-\dfrac{T_{2}}{T_{1}} \quad \therefore \quad 0.3=1-\dfrac{350}{T_{1}}}\) \({\Rightarrow T_{1}=500 {~K}}\) or \({T_{1}=227^{\circ} {C}}\). So correct option is (2)
