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A Carnot engine operates with a large hot reservoir and a much smaller cold reservoir. As a result, the temperature of the hot reservoir remains constant while the temperature of the cold reservoir slowly increases. This temperature change decreases the efficiency of the engine to 0.70 from \(0.75 .\) Find the ratio of the final temperature of the cold reservoir to its initial temperature.

Short Answer

Expert verified
The ratio is 1.2.

Step by step solution

01

Understanding Efficiency of a Carnot Engine

The efficiency \( \eta \) 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. The temperatures \( T_c \) and \( T_h \) must be in Kelvin.
02

Set Efficiency Equations for Initial and Final Conditions

Initially, the efficiency is 0.75. Therefore, \( 0.75 = 1 - \frac{T_{c_1}}{T_h} \). For the final efficiency of 0.70, the equation is \( 0.70 = 1 - \frac{T_{c_2}}{T_h} \).
03

Rearrange Equations to Express Temperatures of the Cold Reservoir

From the equations in Step 2:- For initial conditions, rearrange to \( \frac{T_{c_1}}{T_h} = 0.25 \).- For final conditions, rearrange to \( \frac{T_{c_2}}{T_h} = 0.30 \).
04

Calculate the Ratio of Final to Initial Temperature

Divide the equation for \( T_{c_2} \) by the equation for \( T_{c_1} \):\[\frac{T_{c_2}}{T_{c_1}} = \frac{0.30}{0.25} = 1.2\]
05

Conclude the Result

The ratio of the final temperature \( T_{c_2} \) of the cold reservoir to its initial temperature \( T_{c_1} \) is 1.2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics
Thermodynamics is a fascinating branch of physics that studies the relationships between heat, work, temperature, and energy. It plays a crucial role in understanding how energy changes form and moves between systems. It is particularly important when discussing engines and how they operate.
There are four fundamental laws in thermodynamics, but the majority of our focus, when discussing engines like the Carnot engine, revolves around the first and second laws.
  • The **First Law of Thermodynamics** is essentially the law of energy conservation. It states that energy cannot be created or destroyed; it can only be transferred or changed from one form to another.
  • The **Second Law of Thermodynamics** introduces the concept of entropy, implying that energy systems tend to move towards a state of disorder or entropy. It also declares that heat flows naturally from hot to cold objects.
Understanding these laws is essential for grasping how engines, like the Carnot engine, convert heat into work and function effectively within these principles.
Heat Reservoir
A heat reservoir is a source or destination for heat that provides or absorbs thermal energy at a constant temperature. In the context of a Carnot engine, the two heat reservoirs are the hot and cold reservoirs.
In a Carnot engine:
  • The **hot reservoir** supplies the engine with heat energy, which it converts into work.
  • The **cold reservoir** absorbs or releases heat expelled by the engine after work is done.
The beauty of a Carnot engine is that it operates in a reversible cycle. This means that it is, theoretically, the most efficient way an engine can operate between two temperature limits. However, it's important to note that real-world engines can't achieve the ideal efficiencies of a Carnot cycle due to practical limitations.
Efficiency
Efficiency is a measure of how well a machine, like an engine, converts input energy into useful output work. For a Carnot engine, efficiency is determined by comparing the temperature of the hot and cold reservoirs. This is expressed mathematically as:
\[\eta = 1 - \frac{T_c}{T_h}\]
where \(\eta\) is the efficiency, \(T_c\) is the cold reservoir temperature, and \(T_h\) is the hot reservoir temperature. These temperatures must be in Kelvin, ensuring a positive and consistent scale for measuring efficiency.
Given this formula:
  • A **higher temperature** in the cold reservoir \((T_c)\) leads to a lower efficiency, as seen when the engine's efficiency drops from 0.75 to 0.70 in the problem scenario.
  • A **greater temperature difference** between the hot and cold reservoirs allows for a higher potential efficiency.
Understanding efficiency helps in designing engines and optimizing them for better performance. In real-world terms, it enables engineers to predict how much of the energy will be converted to useful work versus how much is wasted as unusable energy.

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