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Heat \(Q\) flows spontaneously from a reservoir at \(394 \mathrm{K}\) into a reservoir at \(298 \mathrm{K}\). Because of the spontaneous flow, \(2800 \mathrm{J}\) of energy is rendered unavailable for work when a Carnot engine operates between the rescrvoir at \(298 \mathrm{K}\) and a rescrvoir at \(248 \mathrm{K}\). Find \(Q\).

Short Answer

Expert verified
The heat \( Q \) is approximately 3365.75 J.

Step by step solution

01

Understand the Problem

We need to find the heat energy \( Q \) that flows from a hot reservoir at 394 K to a cold reservoir at 298 K given that a Carnot engine renders 2800 J of energy unavailable when operating between two other reservoirs, one at 298 K and the other at 248 K.
02

Calculate Efficiency of Carnot Engine

The efficiency \( \eta \) of a Carnot engine operating between two temperatures \( T_H \) and \( T_C \) is given by \( \eta = 1 - \frac{T_C}{T_H} \). Here, \( T_H = 298 \ \text{K} \) and \( T_C = 248 \ \text{K} \). Substitute these values to find \( \eta \).
03

Calculate Efficiency

Substitute \( T_H = 298 \ \text{K} \) and \( T_C = 248 \ \text{K} \) into the efficiency formula: \[\eta = 1 - \frac{248}{298} = 1 - 0.8322 = 0.1678\] The efficiency of the Carnot engine is 0.1678.
04

Relate Efficiency to Unavailable Work

The efficiency \( \eta \) of the Carnot engine is related to the total work \( W \) and the unavailable energy by \( W = \eta \cdot Q \) and \( Q - W = 2800 \ \text{J} \). We need to find \( Q \).
05

Substitute and Solve for \( Q \)

We know \( W = \eta Q \). Thus, \( Q - \eta Q = 2800 \) implies:\[Q(1 - \eta) = 2800 \]Substitute \( \eta = 0.1678 \):\[Q(1 - 0.1678) = 2800 \]\[Q \cdot 0.8322 = 2800 \]\[Q = \frac{2800}{0.8322} \approx 3365.75 \ \text{J}\]The value of the heat \( Q \) is approximately 3365.75 J.

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It's centered around the concept that heat can be converted into work, and vice versa.
One of the primary topics in thermodynamics is the study of how energy is transferred in systems. These systems can range from simple heat engines to complex biological processes. Descriptions like temperature, pressure, and volume are often used to describe a system's state.
Thermodynamics is essential in understanding how heat engines, like the Carnot engine, operate. Heat engines convert heat into work by taking in energy as heat from a high-temperature reservoir and releasing energy to a lower-temperature reservoir. This process's efficiency is a subject of great interest as it helps us understand limits on conversion capabilities and improve modern energy use.
Heat Transfer
Heat transfer is a fundamental concept in thermodynamics, describing the movement of heat between bodies or within the same body. This movement happens naturally from a higher temperature region to a lower one until equilibrium is reached.
There are three common methods of heat transfer:
  • Conduction: The transfer of heat via direct contact between molecules.
  • Convection: Heat transfer through the movement of fluids or gases.
  • Radiation: The transfer of energy through electromagnetic waves.
In the context of a Carnot engine, heat transfer occurs as the engine absorbs heat from a high-temperature reservoir and expels it to a lower-temperature reservoir. This efficient transfer is crucial because it defines the amount of work the engine can potentially perform. Understanding heat transfer mechanisms helps improve engine designs and optimize energy use in various industrial applications.
Carnot Cycle
The Carnot cycle is an idealized cycle that provides a benchmark for the efficiency of real heat engines. Named after Sadi Carnot, this cycle describes a theoretical engine operating with maximum possible efficiency.
The cycle has four stages, each playing a significant role:
  • Isothermal Expansion: The gas expands while maintaining a constant temperature by absorbing heat from the hot reservoir.
  • Adiabatic Expansion: The gas continues to expand without heat exchange, causing the temperature to fall.
  • Isothermal Compression: Heat is released to the cold reservoir as the gas is compressed at a constant temperature.
  • Adiabatic Compression: Further compression raises the temperature of the gas without heat exchange.
The Carnot cycle illustrates the highest theoretical efficiency attainable by a cycle operating between two temperatures. Though no real engine can achieve Carnot efficiency due to practical limitations, this cycle remains a critical standard for developing more efficient engines.
Energy Efficiency
Energy efficiency in the context of heat engines refers to the engine's ability to convert a specific amount of thermal energy into useful work. For a Carnot engine, energy efficiency is defined precisely by the temperatures of the heat reservoirs. The efficiency formula \[ \eta = 1 - \frac{T_C}{T_H} \]highlights how much of the heat energy absorbed from the hot reservoir can be transformed into work.
In the exercise, the Carnot engine, operating between reservoirs at 298 K and 248 K, has an efficiency of 16.78%. Knowing the efficiency, students can calculate the theoretical maximum work output from an engine given a certain heat input.
Improving energy efficiency has significant implications. It means using less fuel, emitting fewer pollutants, and still achieving the desired output. That's why researchers continuously explore ways to enhance the efficiency of heat engines, making energy consumption more sustainable and cost-effective.

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Most popular questions from this chapter

A Carnot heat pump operates between an outdoor temperature of 265 \(\mathrm{K}\) and an indoor temperature of \(298 \mathrm{K}\). Find its coefficient of performance.

A power plant taps steam superheated by geothermal energy to \(505 \mathrm{K}\) (the temperature of the hot reservoir) and uses the steam to do work in turning the turbine of an electric generator. The steam is then converted back into water in a condenser at \(323 \mathrm{K}\) (the temperature of the cold reservoir), after which the water is pumped back down into the earth where it is heated again. The output power (work per unit time) of the plant is 84000 kilowatts. Determine (a) the maximum efficiency at which this plant can operate and (b) the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

A system does \(4.8 \times 10^{4} \mathrm{J}\) of work, and \(7.6 \times 10^{4} \mathrm{J}\) of heat flows into the system during the process. Find the change in the internal energy of the system.

Heat engines take input energy in the form of heat, use some of that energy to do work, and exhaust the remainder. Similarly, a person can be viewed as a heat engine that takes an input of internal energy, uses some of it to do work, and gives off the rest as heat. Suppose that a trained athlete can function as a heat engine with an efficiency of 0.11. (a) What is the magnitude of the internal energy that the athlete uses in order to do \(5.1 \times 10^{4} \mathrm{J}\) of work? (b) Determine the magnitude of the heat the athlete gives off.

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.

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