/*! 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 10 (a) How much heat transfer occur... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) How much heat transfer occurs to the environment by an electrical power station that uses 1.25×1014J of heat transfer into the engine with an efficiency of 42.0%? (b) What is the ratio of heat transfer to the environment to work output? (c) How much work is done?

Short Answer

Expert verified
Heat transfer to the environment is 7.25×10^{13} J, the ratio of heat transfer to the environment to work output is 1.382, and the work done is 5.25×10^{13} J.

Step by step solution

01

Calculate the work output of the power station

The efficiency (\) of a power station is given by the ratio of work output (W) to heat input (Q_in). In this case, the efficiency is 42.0%, or 0.420, and the heat input is 1.25×10^{14} J. The formula to calculate work output is W = \( \times Q_in. So we calculate W = 0.420 \times 1.25×10^{14} J.
02

Calculate the heat transfer to the environment

The heat transfer to the environment (Q_out) is the difference between the heat input (Q_in) and the work output (W). We already have Q_in = 1.25×10^{14} J and have calculated W in Step 1. Now calculate Q_out = Q_in - W.
03

Find the ratio of heat transfer to the environment to work output

The ratio (R) is found by dividing the heat transfer to the environment (Q_out) by the work output (W). Use the values from the previous steps to calculate R = Q_out / W.
04

Provide the work done by the power station

The work done (W) has already been calculated in Step 1. We can report this as the final result of part (c).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with heat and temperature, and their relation to energy and work. It defines macroscopic variables, such as internal energy, entropy, and pressure, that partly describe a body of matter or radiation. The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system.

The relevance of thermodynamics in our exercise is centered around the concept of heat transfer. In a power station, thermodynamics dictates that the heat energy (\(Q_{in}\)) put into the system is partially converted to work (\(W\)), while the rest is dissipated as waste heat (\(Q_{out}\)) into the surroundings. This phenomenon is governed by the power station's efficiency, which is a measure of how effectively the input heat is converted into useful work. The first law of thermodynamics can be used to calculate the work done by the power station, where \(W = \text{efficiency} \times Q_{in}\), as shown in the step-by-step solution.
Conservation of Energy
The conservation of energy is a fundamental concept that states that the total energy of an isolated system remains constant; energy can be transformed from one form to another but can't be created or lost. This principle is the cornerstone of the first law of thermodynamics and applies to all physical processes, including the operation of a power station.

In our exercise, the electrical power station receives a specific amount of thermal energy (\(Q_{in} = 1.25 \times 10^{14} J\)). The conservation of energy principle allows us to track this energy as it is converted into work output and waste heat. As the efficiency is given as 42%, we understand that only 42% of the input energy is transformed into work (\(W\)), and the remainder must be the heat transferred to the environment (\(Q_{out}\)). This aligns with the provided solution that subtracts the work output from the total heat input to find the heat lost to the environment.
Energy Efficiency
Energy efficiency, in the context of a power station, refers to the ratio of useful work obtained from the energy input. It is a key performance indicator that determines how economically and environmentally favorable a power station is. The higher the efficiency, the more work is produced per unit of energy consumed, and the lower the fuel costs and environmental impact in terms of waste heat released.

From the exercise, the efficiency is 42.0%, indicating that for every 100 joules of heat energy input, only 42 joules are converted into electrical work. The exercise improvement advice focuses on understanding this concept deeply by knowing that the remaining 58 joules become waste heat (\(Q_{out}\)), contributing to the heat transfer to the environment. This not only reflects the conservation of energy but also helps in evaluating the power station's impact on its surroundings. Moreover, the calculation of the work output using the efficiency and heat input is an application of energy efficiency in quantitative analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain how water’s entropy can decrease when it freezes without violating the second law of thermodynamics. Specifically, explain what happens to the entropy of its surroundings.

A rotating space station is said to create "artificial gravity" - a loosely- defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an "artificial gravity" of \(9.80 \mathrm{m} / \mathrm{s}^{2}\) at the rim?

Identify the type of energy transferred to your body in each of the following as either internal energy, heat transfer, or doing work: (a) basking in sunlight; (b) eating food; (c) riding an elevator to a higher floor.

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discovered near its predicted position. But it now appears that the discovery was fortuitous, because Pluto is small and the irregularities in Neptune's orbit were not well known. To illustrate that Pluto has a minor effect on the orbit of Neptune compared with the closest planet to Neptune: (a) Calculate the acceleration due to gravity at Neptune due to Pluto when they are \(4.50 \times 10^{12} \mathrm{m}\) apart, as they are at present. The mass of Pluto is \(1.4 \times 10^{22} \mathrm{kg}\) (b) Calculate the acceleration due to gravity at Neptune due to Uranus, presently about \(2.50 \times 10^{12} \mathrm{m}\) apart, and compare it with that due to Pluto. The mass of Uranus is \(8.62 \times 10^{25} \mathrm{kg}\).

Assume that the turbines at a coal-powered power plant were upgraded, resulting in an improvement in efficiency of 3.32%. Assume that prior to the upgrade the power station had an efficiency of 36% and that the heat transfer into the engine in one day is still the same at 2.50×1014J. (a) How much more electrical energy is produced due to the upgrade? (b) How much less heat transfer occurs to the environment due to the upgrade?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.