Question:
A Carnot engine is operating between a hot body and cold body maintained at temperatures \( T_1 \) and \( T_2 \) respectively. Consider the following three cases:
• Case-I: The temperature of the hot body is changed to \( T_1 + \Delta T \) and cold body is at \( T_2 \).
• Case-II: The temperature of the hot body is at \( T_1 \) and cold body is changed to \( T_2 + \Delta T \).
• Case-III: The temperature of the hot body is at \( T_1 \) and cold body is changed to \( T_2 - \Delta T \). The efficiency of the Carnot cycle is highest for:
A Carnot engine is operating between a hot body and cold body maintained at temperatures \( T_1 \) and \( T_2 \) respectively. Consider the following three cases:
• Case-I: The temperature of the hot body is changed to \( T_1 + \Delta T \) and cold body is at \( T_2 \).
• Case-II: The temperature of the hot body is at \( T_1 \) and cold body is changed to \( T_2 + \Delta T \).
• Case-III: The temperature of the hot body is at \( T_1 \) and cold body is changed to \( T_2 - \Delta T \). The efficiency of the Carnot cycle is highest for:
• Case-I: The temperature of the hot body is changed to \( T_1 + \Delta T \) and cold body is at \( T_2 \).
• Case-II: The temperature of the hot body is at \( T_1 \) and cold body is changed to \( T_2 + \Delta T \).
• Case-III: The temperature of the hot body is at \( T_1 \) and cold body is changed to \( T_2 - \Delta T \). The efficiency of the Carnot cycle is highest for:
Show Hint
Lowering cold reservoir temperature increases efficiency more effectively than raising hot temperature.
Updated On: May 1, 2026
- The efficiency of the Carnot cycle is highest for case-I
- The efficiency of the Carnot cycle is highest for case-II
- The efficiency of the Carnot cycle is highest for case-III
- The efficiency of case-II is higher than case-III
- The efficiency of the Carnot cycle is same for all three cases
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
Carnot efficiency: \[ \eta = 1 - \frac{T_c}{T_h} \]
Step 1: Write efficiencies
\[ \eta_1 = 1 - \frac{T_2}{T_1 + \Delta T} \] \[ \eta_2 = 1 - \frac{T_2 + \Delta T}{T_1} \] \[ \eta_3 = 1 - \frac{T_2 - \Delta T}{T_1} \]
Step 2: Compare physically
Increasing \(T_h\) increases efficiency and decreasing \(T_c\) increases efficiency even more.
Step 3: Conclusion
\[ \eta_3 > \eta_1 > \eta_2 \] \[ \boxed{\text{Efficiency is highest for Case-III}} \]
Carnot efficiency: \[ \eta = 1 - \frac{T_c}{T_h} \]
Step 1: Write efficiencies
\[ \eta_1 = 1 - \frac{T_2}{T_1 + \Delta T} \] \[ \eta_2 = 1 - \frac{T_2 + \Delta T}{T_1} \] \[ \eta_3 = 1 - \frac{T_2 - \Delta T}{T_1} \]
Step 2: Compare physically
Increasing \(T_h\) increases efficiency and decreasing \(T_c\) increases efficiency even more.
Step 3: Conclusion
\[ \eta_3 > \eta_1 > \eta_2 \] \[ \boxed{\text{Efficiency is highest for Case-III}} \]
Was this answer helpful?
0
0
Top KEAM Physics Questions
- A body oscillates with SHM according to the equation (in SI units), $x = 5 cos \left(2\pi t +\frac{\pi}{4}\right) .$ Its instantaneous displacement at $t = 1$ second is
Kepler's second law (law of areas) of planetary motion leads to law of conservation of
- Two capillary tubes A and B of diameter $1\, mm$ and $2\, mm$ respectively are dipped vertically in a liquid. If the capillary rise in A is 6 cm, then the capillary rise in B is
- In the circuit given in figure, 1 and 2 are ammeters. Just after key K is pressed to complete the circuit, the reading will be
- A plate has a length $ 5\pm 0.1\text{ }cm $ and breadth $ 2\pm 0.01\text{ }cm $ . Then the area of the plate is:
View More Questions
Top KEAM carnot cycle Questions
- The Carnot cycle of a reversible heat engine consists of
- A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $0.2$. When $T_2$ is reduced by $50\, K$, its efficiency increases to $0.4$. Then $T_1$ and $T_2$ are respectively
- Efficiency of a Carnot engine
- If the heat supplied to an ideal gas is used up entirely to do work on the surrounding, then the process is called
- A Carnot engine whose low temperature reservoir is at 350 K has an efficiency of 50%. It is desired to increase this to 60%. If the temperature of the low temperature reservoir remains constant, then the temperature of high temperature reservoir must be increased by how many degrees?
View More Questions
Top KEAM Questions
- i.$\quad$ They help in respiration ii.$\quad$ They help in cell wall formation iii.$\quad$ They help in DNA replication iv. $\quad$They increase surface area of plasma membrane Which of the following prokaryotic structures has all the above roles?
- A body oscillates with SHM according to the equation (in SI units), $x = 5 cos \left(2\pi t +\frac{\pi}{4}\right) .$ Its instantaneous displacement at $t = 1$ second is
- The pH of a solution obtained by mixing 60 mL of 0.1 M BaOH solution at 40m of 0.15m HCI solution is
Kepler's second law (law of areas) of planetary motion leads to law of conservation of
- If $\int e^{2x}f' \left(x\right)dx =g \left(x\right)$, then $ \int\left(e^{2x}f\left(x\right) + e^{2x} f' \left(x\right)\right)dx =$
View More Questions