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An engine does 18500 J of work and rejects 6550 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 1091.11 K.

Step by step solution

01

Understand the Problem

We have an engine doing 18500 J of work and rejecting 6550 J of heat into a cold reservoir at 285 K. We need to find the minimum possible temperature of the hot reservoir for this process.
02

Identify the Formula

This problem involves calculating the minimum temperature of the hot reservoir using the concept of Carnot efficiency, which is the maximum efficiency of a heat engine. The Carnot efficiency formula is:\[\eta = 1 - \frac{T_c}{T_h}\]where \(T_c\) is the temperature of the cold reservoir, and \(T_h\) is the temperature of the hot reservoir.
03

Calculate the Efficiency

The efficiency \(\eta\) of the engine can also be calculated using the work done and the heat absorbed by the engine. The formula is:\[\eta = \frac{W}{Q_h}\]where \(W = 18500\, \text{J}\) is the work done and \(Q_h = W + Q_c = 18500 + 6550 = 25050\, \text{J}\) is the total heat absorbed. Therefore, the efficiency is:\[\eta = \frac{18500}{25050} = 0.7389\]
04

Rearrange the Carnot Efficiency Formula

To find the temperature of the hot reservoir \(T_h\), rearrange the Carnot efficiency formula:\[T_h = \frac{T_c}{1 - \eta}\]
05

Input Known Values and Solve for \(T_h\)

Substitute the known values into the rearranged formula. We have \(T_c = 285\, \text{K}\) and \(\eta = 0.7389\):\[T_h = \frac{285}{1 - 0.7389}\]Calculate \(T_h\):\[T_h \approx \frac{285}{0.2611} \approx 1091.11\, \text{K}\]Therefore, the minimum temperature of the hot reservoir is approximately 1091.11 K.

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Key Concepts

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

Heat Engine
A heat engine is a fascinating device that is designed to convert thermal energy into mechanical work. This is a concept often discussed in the study of thermodynamics, an essential branch of physics. In simple terms, a heat engine takes heat from a high-temperature source, does some work, and then releases some heat to a cooler sink.
Here’s how it works:
  • Heat is absorbed from a hot reservoir.
  • The engine does work by utilizing this heat, like powering a car or moving a turbine.
  • Some heat is inevitably expelled to a cold reservoir.
Overall, the key takeaway is that not all the absorbed heat can be converted into work due to inherent inefficiencies. In practical terms, this means that heat engines will always lose some energy, typically through rejected heat to a cooler place.
Thermodynamics
Thermodynamics is the scientific study of energy, heat, and their interactions in systems. It provides the foundational principles that govern heat engines, and one of its key concepts is Carnot efficiency. This is the maximum theoretical efficiency a heat engine can achieve.
The beauty of thermodynamics lies in its laws:
  • The First Law of Thermodynamics states that energy can neither be created nor destroyed. It only changes forms, like converting heat into work.
  • The Second Law of Thermodynamics introduces the concept of inefficiencies and entropy, stating that not all energy can be converted into useful work, leading to inevitable losses in the form of waste heat.
Understanding these laws helps in grasping why engines cannot be 100% efficient and why thermodynamics plays a crucial role in engineering and science.
Temperature of Reservoirs
In the context of heat engines, reservoirs play an integral role. A reservoir is any region that can absorb or supply finite amounts of heat without undergoing a significant change in its temperature. In a heat engine, we deal with two types of reservoirs: the hot reservoir and the cold reservoir.
The significance of their temperatures is vital:
  • The hot reservoir provides the engine with high-temperature heat, which is necessary to do work.
  • The cold reservoir absorbs the residual heat, which is not converted into work, effectively acting as a heat sink.
The efficiency of a heat engine is greatly influenced by the temperatures of these reservoirs. Theoretically, increasing the temperature of the hot reservoir or decreasing the temperature of the cold reservoir will improve the engine's efficiency. This is because efficiency is linked to the temperature difference between the two reservoirs. The greater this difference, the more potential there is for performing work.

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Most popular questions from this chapter

A monatomic ideal gas in a rigid container is heated from \(217 \mathrm{K}\) to \(279 \mathrm{K}\) by adding \(8500 \mathrm{J}\) of heat. How many moles of gas are there in the container?

Due to design changes, the efficiency of an engine increases from 0.23 to \(0.42 .\) For the same input heat \(\left|Q_{\mathrm{H}}\right|,\) these changes increase the work done by the more efficient engine and reduce the amount of heat rejected to the cold reservoir. Find the ratio of the heat rejected to the cold reservoir for the improved engine to that for the original engine.

A refrigerator operates between temperatures of 296 and \(275 \mathrm{K}\). What would be its maximum coefficient of performance?

Even at rest, the human body generates heat. The heat arises because of the body's metabolism - that is, the chemical reactions that are always occurring in the body to generate energy. In rooms designed for use by large groups, adequate ventilation or air conditioning must be provided to remove this heat. Consider a classroom containing 200 students. Assume that the metabolic rate of generating heat is \(130 \mathrm{W}\) for each student and that the heat accumulates during a fifty-minute lecture. In addition, assume that the air has a molar specific heat of \(C_{V}=\frac{5}{2} R\) and that the room (volume \(=1200 \mathrm{m}^{3},\) initial pressure \(=1.01 \times 10^{5} \mathrm{Pa}\), and initial temperature \(=21^{\circ} \mathrm{C}\) ) is sealed shut. If all the heat generated by the students were absorbed by the air, by how much would the air temperature rise during a lecture?

In exercising, a weight lifter loses \(0.150 \mathrm{kg}\) of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is \(1.40 \times\) \(10^{5} \mathrm{J} .\) (a) Assuming that the latent heat of vaporization of perspiration is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg},\) find the change in the internal energy of the weight lifter. (b) Determine the minimum number of nutritional Calories of food (1 nutritional Calorie \(=4186 \mathrm{J}\) ) that must be consumed to replace the loss of internal energy.

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