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(a) What is the work output of a cyclical heat engine having a 22.0% efficiency and 6.00×109J of heat transfer into the engine? (b) How much heat transfer occurs to the environment?

Short Answer

Expert verified
The work output of the engine is 1.32×10^9 J, and the heat transfer to the environment is 4.68×10^9 J.

Step by step solution

01

Understand Work Output

To calculate the work output of a heat engine with a given efficiency, use the formula: Work output = Efficiency × Heat Input. The efficiency should be in decimal form.
02

Convert Efficiency to Decimal

Convert the given efficiency from a percentage to a decimal by dividing by 100. For a 22.0% efficiency, the conversion is: 22.0% / 100 = 0.22.
03

Calculate Work Output

Using the converted efficiency and the given amount of heat input, calculate the work output: Work output = 0.22 × 6.00×10^9 J.
04

Solving for Work Output

Multiply the efficiency in decimal form by the heat transfer into the engine to find the work output: 0.22 × 6.00×10^9 J = 1.32×10^9 J.
05

Understand Heat Transfer to Environment

The heat transfer to the environment is the difference between the heat input and the work output, since the energy not used for work is released as heat to the surroundings.
06

Calculate Heat Transfer to Environment

Subtract the work output from the total heat input to find the heat transfer to the environment: 6.00×10^9 J - 1.32×10^9 J.
07

Solving for Heat Transfer to Environment

Calculate the heat transfer to the environment using the values obtained in step 6: 6.00×10^9 J - 1.32×10^9 J = 4.68×10^9 J.

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Key Concepts

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

Understanding Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In essence, it explores the principles governing the conversion of heat into mechanical work and vice versa. It operates under four fundamental laws that lay the foundations of energy transformations and helps us understand the behavior of systems in response to changes in temperature, volume, and pressure.

For the case of heat engines, which are thermodynamic systems that convert heat energy into mechanical work, these laws prescribe how energy can be efficiently transformed from one form to another. The efficiency of a heat engine, as explored in the textbook exercise, is a critical aspect determined by thermodynamics. It measures the ratio of the useful work output to the heat input into the system. Efficiency is intrinsic to thermodynamic cycles and is a key factor in evaluating the performance of real-world systems such as car engines and power plants.
Calculating Work Output
Work output is a term frequently used in engineering and physics that refers to the energy transferred by a system during a process where force is applied. To calculate the work output of a heat engine, it is crucial to know both the amount of heat energy that enters the system and the efficiency of the engine in converting that heat into work.

In the textbook exercise, the formula to find the work output is given by multiplying the efficiency (in decimal form) with the heat input. This mathematical approach translates the conceptual understanding of energy conversion into a quantifiable measure. It's vital to apply the correct units for energy—joules (J) in the International System of Units—and to ensure that efficiency is converted from a percentage to a decimal before calculations.

Tip for Calculation:

Always double-check the units and conversion of efficiency to avoid common mistakes. Remember, the more accurate your input values, the more reliable your calculated work output will be.
Heat Transfer Dynamics
Heat transfer is another fundamental concept in thermodynamics, defining how thermal energy moves from one body or system to another with a different temperature. In a heat engine, some of the energy is always lost to the environment because no engine can be 100% efficient due to the second law of thermodynamics. Thus, any energy that is not converted into work is ultimately released as heat.

The textbook exercise demonstrates calculating the amount of heat transferred to the environment by subtracting the work output from the total heat input. This ensures the principle of conservation of energy is respected, where the total energy remains constant in an isolated system, merely transitioning from one form to another or from one part of the system to another.

Note on Energy Conservation:

Determining the heat transfer to the environment is not just a measure of energy 'wasted' but is a critical part in calculating thermal efficiency and designing systems to minimize energy loss for environmental and economic benefits.

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Most popular questions from this chapter

Modern roller coasters have vertical loops like the one shown in Figure \(6.38 .\) The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \(15.0 \mathrm{m}\) and the downward acceleration of the car is \(1.50 \mathrm{g} ?\) Figure 6.38 Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than \(g\) so that the passengers do not lose contact with their seats nor do they need seat belts to keep them in place.

(a) Calculate Earth's mass given the acceleration due to gravity at the North Pole is \(9.830 \mathrm{m} / \mathrm{s}^{2}\) and the radius of the Earth is \(6371 \mathrm{km}\) from center to pole. (b) Compare this with the accepted value of \(5.979 \times 10^{24} \mathrm{kg}\).

Give an explanation of how food energy (calories) can be viewed as molecular potential energy (consistent with the atomic and molecular definition of internal energy).

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The temperature of a rapidly expanding gas decreases. Explain why in terms of the first law of thermodynamics. (Hint: Consider whether the gas does work and whether heat transfer occurs rapidly into the gas through conduction.)

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