/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 An engine does 18500 \(\mathrm{J... [FREE SOLUTION] | 91Ó°ÊÓ

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An engine does 18500 \(\mathrm{J}\) of work and rejects 6550 \(\mathrm{J}\) of heat into a cold reservoir whose temperature is 285 \(\mathrm{K}\) . What would be the smallest possible temperature of the hot reservoir?

Short Answer

Expert verified
The smallest possible temperature of the hot reservoir is approximately 1087.79 K.

Step by step solution

01

Understand the Concepts

In this problem, we are dealing with the efficiency of a heat engine. The efficiency formula for a heat engine is given by \[ e = \frac{W}{Q_h} \]where \( W \) is the work done by the engine and \( Q_h \) is the heat absorbed by the engine. The Carnot efficiency for an ideal engine is \[ e_c = 1 - \frac{T_c}{T_h} \]where \( T_c \) and \( T_h \) are the temperatures of the cold and hot reservoirs, respectively.
02

Calculate Heat Absorbed

To find the heat absorbed \( Q_h \), use the formula for the work-energy principle in a heat engine: \[ Q_h = W + Q_c = 18500 \ \mathrm{J} + 6550 \ \mathrm{J} = 25050 \ \mathrm{J} \]where \( Q_c \) is the heat rejected into the cold reservoir.
03

Calculate Actual Efficiency

Use the formula for efficiency \( e \):\[ e = \frac{W}{Q_h} = \frac{18500}{25050} \approx 0.738 \]This is the efficiency of the actual engine.
04

Solve for Minimum Hot Reservoir Temperature

To find the minimum temperature of the hot reservoir that achieves this efficiency:Use the Carnot efficiency formula:\[ 1 - \frac{T_c}{T_h} = e \]Substituting the known values:\[ 1 - \frac{285}{T_h} = 0.738 \]Solve for \( T_h \):\[ 0.262 = \frac{285}{T_h} \implies T_h = \frac{285}{0.262} \approx 1087.79 \ \mathrm{K} \]
05

State the Conclusion

Thus, the smallest possible temperature of the hot reservoir, based on Carnot efficiency principles, is \( T_h \approx 1087.79 \ \mathrm{K} \).

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Key Concepts

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

Carnot efficiency
Carnot efficiency is a crucial concept in understanding how efficient a heat engine can theoretically become. This efficiency is named after Sadi Carnot, a French physicist who first described it. The Carnot efficiency provides an upper limit to the efficiency that any heat engine can achieve, regardless of the working substance or the cycle of operations. It encapsulates the fundamental laws of thermodynamics that govern energy transformations.

Mathematically, the Carnot efficiency is expressed as:
  • \( e_c = 1 - \frac{T_c}{T_h} \)
Here, \( T_c \) and \( T_h \) are the absolute temperatures of the cold and hot reservoirs, respectively, measured in kelvins. This formula tells us that the efficiency depends solely on the temperatures of these two reservoirs.

The closer the temperature of the hot reservoir \( T_h \) is to infinity, and the cold reservoir \( T_c \) is to zero, the closer the Carnot efficiency approaches 100%. However, it's important to realize that a 100% efficient engine is practically impossible due to physical constraints, highlighting the theoretical nature of Carnot's principles.
Thermodynamics
Thermodynamics is the study of energy transformations and its principles form the basis for understanding heat engines like the one in our example. It involves concepts such as heat, work, internal energy, and how these are converted into each other. The core laws of thermodynamics are particularly relevant when analyzing the operations of any heat engine.

The First Law of Thermodynamics, often known as the conservation of energy, states that the energy in a closed system is constant.
  • This means that energy can be transformed from one form to another but cannot be created or destroyed.
In the context of a heat engine, this law is vital as it represents the relationship among the work done by the engine, the heat input from the hot reservoir, and the heat rejected to the cold reservoir.

The Second Law of Thermodynamics is equally critical and introduces the concept of entropy. It dictates that energy transfers are not 100% efficient and that entropy, a measure of randomness or disorder, tends to increase over time. This principle underlines the reason why even the best-designed machines cannot be perfectly efficient.
Temperature of hot and cold reservoirs
The temperature of the hot and cold reservoirs is essential in determining the efficiency of any heat engine. These temperatures directly influence the efficiency through the Carnot and real efficiency calculations.
  • The higher the temperature difference between these two reservoirs, the more potential there is for the engine to convert heat into work.
In our problem, the cold reservoir is at a temperature of 285 K. This sets one part of the boundary condition for calculating the engine's efficiency.

To determine the smallest possible temperature for the hot reservoir \( T_h \) in this case, we utilize the Carnot efficiency formula:
  • Substituting the efficiency we calculated, we rearrange to solve for \( T_h \) by isolating it on one side of the equation.
Ultimately, these temperatures are crucial because they not only help dictate the machine's possible efficiency but also intrinsically limit how much work can be extracted from a given amount of heat energy.
Heat work calculation
Understanding heat work calculation is key to comprehending how heat engines operate and how efficient they can be. In simple terms, a heat engine is a device that converts heat energy into mechanical work.
  • The efficiency of this process can be calculated by the ratio of the work output \( W \) to the heat absorbed from the hot reservoir \( Q_h \).
In our example, the engine does 18500 J of work and rejects 6550 J of heat to the cold reservoir. To find \( Q_h \), the total heat input into the engine, we sum the work done and the heat expelled:
  • \( Q_h = 18500 \ \text{J} + 6550 \ \text{J} = 25050 \ \text{J} \)
This calculation is pivotal in determining the actual efficiency of the engine, defined as:
  • \( e = \frac{W}{Q_h} \)
With these calculations, we not only estimate the real-world efficiency but also uncover the theoretical limitations outlined by the laws of thermodynamics and Carnot's insights.

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Most popular questions from this chapter

Three moles of a monatomic ideal gas are heated at a constant volume of 1.50 \(\mathrm{m}^{3}\) . The amount of heat added is \(5.24 \times 10^{3} \mathrm{J}\) . ( a) What is the change in the temperature of the gas? (b) Find the change in its internal energy. (c) Determine the change in pressure.

The temperature of 2.5 mol of a monatomic ideal gas is 350 \(\mathrm{K}\) . The internal energy of this gas is doubled by the addition of heat. How much heat is needed when it is added at \((\mathrm{a})\) constant volume and \((\mathrm{b})\) constant pressure?

The sun is a sphere with a radius of \(6.96 \times 10^{8} \mathrm{m}\) and an average surface temperature of 5800 k. Determine the amount by which the sun's thermal radiation increases the entropy of the entire universe each second. Assume that the sun is a perfect blackbody, and that the average temperature of the rest of the universe is 2.73 \(\mathrm{K}\) . Do not consider the thermal radiation absorbed by the sun from the rest of the universe.

A Carnot engine operates with an efficiency of 27.0\(\%\) when the temperature of its cold reservoir is 275 \(\mathrm{K}\) . Assuming that the temperature of the hot reservoir remains the same, what must be the temperature of the cold reservoir in order to increase the efficiency to 32.0\(\%\) ?

The drawing (not to scale) shows the way in which the pressure and volume change for an ideal gas that is used as the working substance in a Carnot engine. The gas begins at point a (pressure \(=P_{a},\) volume \(=V_{\mathrm{a}}\) and expands isothermally at temperature \(T_{\mathrm{H}}\) until point b (pressure \(=P_{\mathrm{b}}\) , volume \(=V_{\mathrm{b}}\) is reached. During this expansion, the input heat of magnitude \(\left|Q_{\mathrm{H}}\right|\) enters the gas from the hot reservoir of the engine. Then, from point b to point \(c\) (pressure \(=P_{c},\) volume \(=V_{c} )\) the gas expands adiabatically. Next, the gas is compressed isothermally at temperature \(T_{\mathrm{C}}\) from point \(c\) to point \(\mathrm{d}\) (pressure \(=P_{\mathrm{d}},\) volume \(=V_{\mathrm{d}} )\) During this compression, heat of magnitude \(\left|Q_{\mathrm{C}}\right|\) is rejected to the cold reservoir of the engine. Finally, the gas is compressed adiabatically from point d to point a, where the gas is back in its initial state. The overall process a to b to c to d to a is called a Carnot cycle. Prove for this cycle that \(\left|Q_{\mathrm{c}}\right| /\left|Q_{\mathrm{H}}\right|=T_{\mathrm{C}} / T_{\mathrm{H}} .\)

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