What is the minimum amount of work that must be done to extract \(500.0 \mathrm{~J}\) of heat from a massive object at a temperature of \(27.0^{\circ} \mathrm{C}\) while releasing heat to a high temperature reservoir with a temperature of \(100.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The minimum amount of work required is 2554.0 J.

Step by step solution

01

Convert the given temperatures to Kelvin

First, we need to convert both the low and high temperature reservoirs from Celsius to Kelvin. To do that, add 273.15 to each given temperature: \(T_{1} \text{(Kelvin)} = 27.0 + 273.15 = 300.15\,\mathrm{K}\) \(T_{2} \text{(Kelvin)} = 100.0 + 273.15 = 373.15\,\mathrm{K}\)
02

Calculate the efficiency of the heat engine

In order to calculate the minimum amount of work required, we need to find the efficiency of the heat engine, which is determined by the formula: \(\eta = 1 - \frac{T_{1}}{T_{2}}\) Plugging in the values for \(T_{1}\) and \(T_{2}\), we get: \(\eta = 1 - \frac{300.15\,\mathrm{K}}{373.15\,\mathrm{K}} = 1 - 0.8043 = 0.1957\)
03

Calculate the work done using the efficiency

To find the work done, we'll use the efficiency formula rearranged as follows: \(W = \frac{Q_{1}}{\eta}\) We are given that \(Q_{1}\), the heat extracted from the massive object, is \(500.0\,\mathrm{J}\). So, plugging in the efficiency value and the given heat extraction, we get: \(W = \frac{500.0\,\mathrm{J}}{0.1957} = 2554.0\,\mathrm{J}\) Therefore, the minimum amount of work that must be done to extract \(500.0\,\mathrm{J}\) of heat from the massive object is \(2554.0\,\mathrm{J}\).

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Key Concepts

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

Heat Engine Efficiency
Understanding the efficiency of a heat engine is crucial when delving into thermodynamic problems. The efficiency, denoted by \(\eta\), represents how well an engine converts heat into work. It is a ratio of the work output to the heat input, often expressed as a percentage. For an ideal heat engine operating between two temperatures, the efficiency is given by the formula \(\eta = 1 - \frac{T_1}{T_2}\), where \(T_1\) is the temperature of the cold reservoir and \(T_2\) is the temperature of the hot reservoir, both in Kelvin.

In our exercise, the efficiency calculation reveals the theoretical limit of performance for the heat engine in question. Since real engines cannot be 100% efficient due to unavoidable heat losses and other irreversibilities, the calculated efficiency serves as an upper benchmark. The exercise demonstration helps solidify the concept that the efficiency depends on the temperature difference: the greater the difference, the more efficient the engine can potentially be.
Temperature Conversion
Temperature conversion is a fundamental step in thermodynamics, which is the science dealing with heat and temperature, and their relation to energy and work. Since most thermodynamic formulas require temperature in Kelvin, it's essential to convert Celsius or Fahrenheit to Kelvin whenever necessary. The conversion from Celsius to Kelvin is straightforward: simply add 273.15 to the Celsius temperature to obtain the Kelvin equivalent. Therefore, \(27.0^\circ\mathrm{C}\) becomes \(300.15\mathrm{K}\), and \(100.0^\circ\mathrm{C}\) becomes \(373.15\mathrm{K}\).

The elegance of the Kelvin scale lies in its direct relation to absolute zero, the lowest possible temperature where classical thermal motion ceases, making it an ideal choice for scientific calculations.
Kelvin Scale
The Kelvin scale is a thermodynamic temperature scale based on absolute zero, the point at which particles have minimum thermal motion, theoretically. This scale is crucial in the scientific community because it provides a standard for temperature measurements in thermodynamic processes. Unlike Celsius or Fahrenheit, Kelvin is not measured in degrees; instead, it uses 'Kelvin' (K) as the unit. The lack of negative numbers on this scale simplifies many physical laws, making it indispensable in thermodynamic calculations, such as in our heat engine efficiency problem.
Heat Transfer
Heat transfer is a fundamental concept in thermodynamics, representing the movement of thermal energy from one thing to another due to a temperature difference. There are three main modes of heat transfer: conduction, convection, and radiation. In the context of our exercise, the heat engine operates by transferring heat from a hot reservoir to a cold one and, in the process, doing work.

This heat transfer concept connects closely with the engine's efficiency and the Kelvin scale temperatures of the reservoirs. Understanding how heat flows from high to lower temperatures without external work is key to grasping the logic behind the second law of thermodynamics, which implies no engine can be entirely efficient, as some energy will always transfer to the surroundings as waste heat.

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Most popular questions from this chapter

Other state variables useful for characterizing different classes of processes can be defined from \(E_{\text {int }}, S, P\), and \(V\). These include the enthalpy, \(H=E_{\text {int }}+p V\), the Helmholtz free energy, \(A=E_{\text {int }}-T S,\) and the Gibbs free energy, \(G=E_{\text {int }}+p V-T S\). a) Write the differential equations for \(d H, d A,\) and \(d G\). b) All of these are also exact differentials. What relationships follow from this fact? Use the First Law to simplify.

Consider a Carnot engine that works between thermal reservoirs with temperatures of \(1000.0 \mathrm{~K}\) and \(300.0 \mathrm{~K}\). The average power of the engine is \(1.00 \mathrm{~kJ}\) per cycle. a) What is the efficiency of this engine? b) How much energy is extracted from the warmer reservoir per cycle? c) How much energy is delivered to the cooler reservoir?

Suppose a person metabolizes \(2000 .\) kcal/day. a) With a core body temperature of \(37.0^{\circ} \mathrm{C}\) and an ambient temperature of \(20.0^{\circ} \mathrm{C}\), what is the maximum (Carnot) efficiency with which the person can perform work? b) If the person could work with that efficiency, at what rate, in watts, would they have to shed waste heat to the surroundings? c) With a skin area of \(1.50 \mathrm{~m}^{2}\), a skin temperature of \(27.0^{\circ} \mathrm{C}\) and an effective emissivity of \(e=0.600,\) at what net rate does this person radiate heat to the \(20.0^{\circ} \mathrm{C}\) surroundings? d) The rest of the waste heat must be removed by evaporating water, either as perspiration or from the lungs At body temperature, the latent heat of vaporization of water is \(575 \mathrm{cal} / \mathrm{g}\). At what rate, in grams per hour, does this person lose water? e) Estimate the rate at which the person gains entropy. Assume that all the required evaporation of water takes place in the lungs, at the core body temperature of \(37.0^{\circ} \mathrm{C}\).

Which of the following processes (all constanttemperature expansions) produces the most work? a) An ideal gas consisting of 1 mole of argon at \(20^{\circ} \mathrm{C}\) expands from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\). b) An ideal gas consisting of 1 mole of argon at \(20^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\). c) An ideal gas consisting of 2 moles of argon at \(10^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\). d) An ideal gas consisting of 1 mole of argon at \(40^{\circ} \mathrm{C}\) expands from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\) e) An ideal gas consisting of 1 mole of argon at \(40^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\).

Is it a violation of the Second Law of Thermodynamics to capture all the exhaust heat from a steam engine and funnel it back into the system to do work? Why or why not?

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