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In a Carnot engine the hot reservoir is \(72.0 \mathrm{C}^{\circ}\) warmer than the cold reservoir. The engine's efficiency is \(12.5 \% .\) What are the Kelvin temperatures of the two reservoirs?

Short Answer

Expert verified
The Kelvin temperatures of the hot and cold reservoirs are \(576 K\) and \(504 K\) respectively.

Step by step solution

01

Title

Convert the temperature difference from Celsius to Kelvin: \(\Delta T_{celsius} = \Delta T_{kelvin}\) because the size of degree Celsius and degree Kelvin are the same, only their zeroes are at different levels. So, \(\Delta T_{kelvin} = 72 K\).
02

Title

Write down the equation of Carnot engine's efficiency \(eff = 1 - \frac{T_{cold}}{T_{hot}}\) where \(eff\) is the efficiency of the Carnot engine and rearrange the equation for \(T_{cold}\): \(T_{cold} = T_{hot}(1 - eff)\).
03

Title

Substitute the known values of efficiency and temperature difference into this rearranged equation: \(T_{hot} = \frac{\Delta T_{kelvin}}{eff} = \frac{72K}{0.125} = 576 K\).
04

Title

Substitute \(T_{hot}\) to \(T_{cold} = T_{hot}(1 - eff)\) to find the cold temperature: \(T_{cold} = 576 K(1 - 0.125) = 504 K\).

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Key Concepts

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

Efficiency of a Carnot Engine
The efficiency of a Carnot engine is a key concept in thermodynamics. It tells us how much useful work can be extracted from a heat engine relative to the heat energy put into the system. The Carnot efficiency is the formula: \[ eff = 1 - \frac{T_{cold}}{T_{hot}} \]where:
  • \( eff \) is the efficiency.
  • \( T_{cold} \) is the temperature of the cold reservoir in Kelvin.
  • \( T_{hot} \) is the temperature of the hot reservoir in Kelvin.
This efficiency is theoretical, representing the maximum possible efficiency a engine can achieve under ideal conditions. All real engines have efficiencies less than the Carnot limit due to irreversibilities and non-idealities. Understanding efficiency helps us in making engines more effective and economical by minimizing wasted energy.
Kelvin Temperature
The Kelvin scale is an absolute temperature measurement scale often used in scientific calculations. Unlike degrees Celsius which can have negative values, the Kelvin scale starts at absolute zero, the coldest possible temperature. This makes it ideal for use in thermodynamics, including calculations involving heat engines like Carnot engines.Temperature differences remain the same between Celsius and Kelvin, but actual temperature values differ:
  • To convert from Celsius to Kelvin, use: \[ T(K) = T(°C) + 273.15 \]
Absolute temperature impacts the efficiency of engineering systems. The higher the hot reservoir's temperature in Kelvin, the greater the potential efficiency. Therefore, converting all temperatures into Kelvin when solving thermodynamics problems is crucial.
Heat Reservoirs in a Carnot Engine
Heat reservoirs are fundamental in understanding how Carnot engines operate. They are large bodies that supply or absorb finite amounts of heat without changing their own temperature significantly. The efficiency of a Carnot engine greatly depends on the temperatures of these reservoirs. In a theoretical Carnot cycle:
  • The hot reservoir supplies heat \( Q_h \) at temperature \( T_{hot} \).
  • The cold reservoir absorbs heat \( Q_c \) at temperature \( T_{cold} \).
The temperature difference between these two reservoirs is critical for engine performance. The greater the temperature difference, the higher the engine's potential efficiency. For the Carnot cycle to achieve maximum efficiency, these reservoirs are assumed to be perfect (maintain constant temperatures) and infinite. Understanding these concepts helps explain the limits of real-world engines and potential improvements.

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Most popular questions from this chapter

A Carnot engine performs \(2.5 \times 10^{4} \mathrm{~J}\) of work in each cycle and has an cfficicncy of \(66 \%\). (a) How much heat docs the cngine extract from its heat source in each cycle? (b) If the engine exhausts heat at room temperature \(\left(20.0^{\circ} \mathrm{C}\right),\) what is the temperature of its heat source?

A sophomore with nothing bctter to do adds heat to \(0.350 \mathrm{~kg}\) of ice at \(0.0^{\circ} \mathrm{C}\) until it is all melted. (a) What is the change in entropy of the waler? (b) The source of heat is a very massive object at \(25.0^{\circ} \mathrm{C}\). What is the change in entropy of this object? (c) What is the total change in entropy of the water and the heat source?

An ice-making machine operates in a Carnot cycle. It takes heat from water at \(0.0^{\circ} \mathrm{C}\) and rejects heat to a room at \(24.0^{\circ} \mathrm{C}\). Suppose that \(85.0 \mathrm{~kg}\) of water at \(0.0^{\circ} \mathrm{C}\) are converted to ice at \(0.0^{\circ} \mathrm{C}\). (a) How much heat is discharged into the room? (b) How much energy must be supplied to the device?

A gasoline engine has a power output of \(180 \mathrm{~kW}\) (about \(241 \mathrm{hp}\) ). Its thermal efliciency is \(28.0 \%\). (a) How much heat must be supplied to the engine per second? (b) How much heat is discarded by the engine per second?

Three moles of an ideal gas undergo a reversible isothermal compression at \(20.0^{\circ} \mathrm{C}\). During this compression, \(1850 \mathrm{~J}\) of work is done on the gas. What is the change of cntropy of the gas?

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