A Carnot engine operating between two reservoirs has efficiency \(\frac{1}{3}\) When the temperature of cold reservoir raised by \(x\), its efficiency decreases to \(\frac{1}{6}\) The value of \(x\), if the temperature of hot reservoir is \(99^{\circ} C\), will be :
Show Hint
Remember to convert temperatures to Kelvin when working with Carnot engine problems. The efficiency formula relates the temperatures of the hot and cold reservoirs to the engine’s efficiency.
- $62 K$
- $16.5 I$
- $33 K$
- $66 K$
The Correct Option is A
Solution and Explanation
Step 1: Convert Celsius to Kelvin
The temperature of the hot reservoir is given as \( T_H = 99^\circ C \). Convert this to Kelvin:
\[ T_H = 99 + 273 = 372 \, \text{K} \]
Step 2: Use the Carnot Efficiency Formula
The efficiency of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_C}{T_H} \]
where \( T_C \) is the temperature of the cold reservoir and \( T_H \) is the temperature of the hot reservoir.
Step 3: Calculate the Initial Cold Reservoir Temperature
Initially, the efficiency is \( \frac{1}{3} \). So,
\[ \frac{1}{3} = 1 - \frac{T_C}{372} \]
\[ \frac{T_C}{372} = 1 - \frac{1}{3} = \frac{2}{3} \]
\[ T_C = \frac{2}{3} \times 372 = 248 \, \text{K} \]
Step 4: Calculate the Cold Reservoir Temperature After the Increase
When the cold reservoir temperature is increased by \( x \), the new temperature is \( T_C + x \), and the efficiency becomes \( \frac{1}{6} \). So,
\[ \frac{1}{6} = 1 - \frac{T_C + x}{T_H} \]
\[ \frac{T_C + x}{T_H} = 1 - \frac{1}{6} = \frac{5}{6} \]
\[ T_C + x = \frac{5}{6} \times 372 \]
\[ 248 + x = \frac{5}{6} \times 372 \]
Step 5: Solve for \( x \)
\[ 248 + x = 310 \]
\[ x = 310 - 248 = 62 \, \text{K} \]
Conclusion: The value of \( x \) is 62 K (Option 2).
Top Questions on carnot cycle
- A Carnot engine operates between temperatures of $ 600 \, \text{K} $ and $ 300 \, \text{K} $. If it absorbs $ 900 \, \text{J} $ of heat from the source, how much work does it perform?
- BITSAT - 2025
- Physics
- carnot cycle
- A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Calculate the efficiency of the engine.
- AP PGECET - 2024
- Nano Technology
- carnot cycle
- Identify the equation typically used to describe the efficiency of a cyclic process.
- AP PGECET - 2024
- Nano Technology
- carnot cycle
- The P-V diagram of a Carnot's engine is shown in the graph below. The engine uses 1 mole of an ideal gas as working substance. From the graph, the area enclosed by the P-V diagram is
[ The heat supplied to the gas is 8000 J ]
- KCET - 2023
- Physics
- carnot cycle
- In a Carnot’s engine, as the gas gives heat energy to the sink, then the temperature of the sink
- AP EAPCET - 2023
- Physics
- carnot cycle
Questions Asked in JEE Main exam
Statement-1: \( \text{ClF}_3 \) has 3 possible structures.
Statement-2: \( \text{III} \) is the most stable structure due to least lone pair-bond pair (lp-bp) repulsion.
Which of the following options is correct?
- JEE Main - 2025
- Molecular Stability
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \), where \( \omega \) is the frequency of the supply across resistor \( R \) and inductor \( L \). If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
In light of the above statements, choose the most appropriate answer from the options given below:
- JEE Main - 2025
- Electromagnetism
- Let \( (2, 3) \) be the largest open interval in which the function \( f(x) = 2 \log_e (x - 2) - x^2 + ax + 1 \) is strictly increasing, and \( (b, c) \) be the largest open interval, in which the function \( g(x) = (x - 1)^3 (x + 2 - a)^2 \) is strictly decreasing. Then \( 100(a + b - c) \) is equal to:
- JEE Main - 2025
- Quadratic Equations
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Complex numbers
- If \( \alpha \) and \( \beta \) are the roots of the equation \( 2z^2 - 3z - 2i = 0 \), where \( i = \sqrt{-1} \), then \[ 16 \cdot {Re} \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \cdot {Im} \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \] is equal to:
- JEE Main - 2025
- Complex numbers
Concepts Used:
Thermodynamics
Thermodynamics in physics is a branch that deals with heat, work and temperature, and their relation to energy, radiation and physical properties of matter.
Important Terms
System
A thermodynamic system is a specific portion of matter with a definite boundary on which our attention is focused. The system boundary may be real or imaginary, fixed or deformable.
There are three types of systems:
- Isolated System – An isolated system cannot exchange both energy and mass with its surroundings. The universe is considered an isolated system.
- Closed System – Across the boundary of the closed system, the transfer of energy takes place but the transfer of mass doesn’t take place. Refrigerators and compression of gas in the piston-cylinder assembly are examples of closed systems.
- Open System – In an open system, the mass and energy both may be transferred between the system and surroundings. A steam turbine is an example of an open system.
Thermodynamic Process
A system undergoes a thermodynamic process when there is some energetic change within the system that is associated with changes in pressure, volume and internal energy.
There are four types of thermodynamic process that have their unique properties, and they are:
- Adiabatic Process – A process in which no heat transfer takes place.
- Isochoric Process – A thermodynamic process taking place at constant volume is known as the isochoric process.
- Isobaric Process – A process in which no change in pressure occurs.
- Isothermal Process – A process in which no change in temperature occurs.
Laws of Thermodynamics
Zeroth Law of Thermodynamics
The Zeroth law of thermodynamics states that if two bodies are individually in equilibrium with a separate third body, then the first two bodies are also in thermal equilibrium with each other.
First Law of Thermodynamics
The First law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic processes, distinguishing three kinds of transfer of energy, as heat, as thermodynamic work, and as energy associated with matter transfer, and relating them to a function of a body's state, called internal energy.
Second Law of Thermodynamics
The Second law of thermodynamics is a physical law of thermodynamics about heat and loss in its conversion.
Third Law of Thermodynamics
Third law of thermodynamics states, regarding the properties of closed systems in thermodynamic equilibrium: The entropy of a system approaches a constant value when its temperature approaches absolute zero.