Question

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

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

Step-by-step solution

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: \(T_{1} \text{(Kelvin)} = 27.0 + 273.15 = 300.15\,\mathrm{K}\) \(T_{2} \text{(Kelvin)} = 100.0 + 273.15 = 373.15\,\mathrm{K}\)

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: \(\eta = 1 - \frac{300.15\,\mathrm{K}}{373.15\,\mathrm{K}} = 1 - 0.8043 = 0.1957\)

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 \(2554.0\,\mathrm{J}\).

Key concepts

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.