/*! 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 62 It has been suggested that a hea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

It has been suggested that a heat engine could be developed that made use of the temperature difference between water at the surface of the ocean and water several hundred meters deep. In the tropics, the temperatures may be \(27^{\circ} \mathrm{C}\) and \(4^{\circ} \mathrm{C}\) respectively. (a) What is the maximum efficiency such an engine could have? (b) Why might such an engine be feasible in spite of the low efficiency? (c) Can you imagine any adverse environmental effects that might occur?

Short Answer

Expert verified
(a) Maximum efficiency is 7.66%. (b) Feasibility arises from a renewable resource. (c) Environmental effects may include ecological disruptions.

Step by step solution

01

Understanding the Problem

First, understand that we need to calculate the maximum efficiency of a heat engine operating between two temperature reservoirs. This efficiency is derived from the Carnot efficiency formula which is used for an ideal engine working reversibly between two reservoirs.
02

Convert Temperatures to Kelvin

To use the Carnot efficiency formula, temperatures must be in Kelvin. Convert the given temperatures from Celsius to Kelvin: \(T_1 = 27 + 273.15 = 300.15\ K\) and \(T_2 = 4 + 273.15 = 277.15\ K\).
03

Use the Carnot Efficiency Formula

The Carnot efficiency \(\eta\) is calculated as:\[\eta = 1 - \frac{T_2}{T_1}\]Substitute the Kelvin temperatures you found into the formula:\[\eta = 1 - \frac{277.15}{300.15} = 0.0766\]This gives a maximum efficiency of 7.66%.
04

Assess Feasibility

Though the efficiency is low, such engines might be feasible because the energy source (ocean temperature difference) is vast and renewable, which can compensate for low efficiency in large-scale applications.
05

Consider Environmental Effects

Operating such an engine on a large scale could disrupt local marine ecology by changing temperature distributions, affecting marine life and nutrient flows.

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.

Heat Engine
Heat engines are devices that convert thermal energy into mechanical energy by exploiting the heat flow between a hot and a cold reservoir. They operate based on fundamental thermodynamic principles.
In the context of an ocean thermal energy application, the hot reservoir is the warm surface water, while the cold reservoir is the cooler deep ocean water.
The temperature difference between these two water layers allows the heat engine to do work.
Heat engines always require a temperature difference to operate, without it, work cannot be extracted from the system.
  • Thermal energy transfers from the hot reservoir to the cold reservoir.
  • This transfer results in the engine performing work.
  • Efficiency is a measure of how well the energy is converted to work.
Understanding these basics helps in grasping how such an engine harnesses energy from ocean temperature gradients.
Temperature Difference
The temperature difference in a heat engine is crucial for its operation and efficiency. It represents the potential energy available for conversion into work.
In the case presented, the surface temperature of the ocean is given as 27 degrees Celsius, while the deeper ocean temperature is 4 degrees Celsius.
These differences must also be converted into Kelvin for thermodynamic calculations. This gives us:
  • Surface Temperature: \(T_1 = 300.15 \ K\)
  • Deep Ocean Temperature: \(T_2 = 277.15 \ K\)
The greater the temperature difference, the more potential there is for efficient energy conversion.
In this specific example, although the temperature difference is relatively small, it is still capable of driving a heat engine due to the abundance of ocean water acting as a source and sink.
Ocean Thermal Energy
Ocean thermal energy utilizes the temperature difference in ocean water to generate energy. This energy source is both vast and renewable, making it a promising, sustainable energy option.
The tropics provide an ideal location for such a system since surface temperatures are consistently higher than deeper waters.
  • Renewable energy source: Unlike conventional fossil fuels, ocean thermal energy is sustainable.
  • Availability: With the vastness of oceans, this energy can be harnessed on a large scale.
  • Consistency: Ocean temperatures are fairly stable, providing a consistent energy source.
Despite the low efficiency of such engines, the renewable nature and abundance of the energy source make it feasible on a large scale.
Additionally, with advanced technology, the minor energy difference can be utilized more effectively.
Efficiency Calculation
Efficiency calculation in thermodynamics is essential to determine how well a system converts energy. The Carnot efficiency formula is generally used to find the theoretical maximum efficiency of a reversible heat engine operating between two thermal reservoirs.
It is expressed as:\[\eta = 1 - \frac{T_2}{T_1}\]Where:
  • \(\eta\) is the efficiency.
  • \(T_1\) is the temperature of the hot reservoir (in Kelvin).
  • \(T_2\) is the temperature of the cold reservoir (in Kelvin).
In the problem example, substituting the given temperatures into the formula gives:
\[\eta = 1 - \frac{277.15}{300.15} = 0.0766\]This results in a maximum theoretical efficiency of 7.66%.
Understanding efficiency helps gauge how effectively the engine can convert heat from the ocean into mechanical work. However, practicality in energy conversion must also consider real-world factors such as losses and environmental impacts aside from theoretical efficiency.

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

Refrigeration units can be rated in "tons." A 1 -ton air conditioning system can remove sufficient energy to freeze 1 British ton \((2000\) pounds \(=909 \mathrm{kg})\) of \(0^{\circ} \mathrm{C}\) water into \(0^{\circ} \mathrm{C}\) ice in one 24 -h day. If, on a \(35^{\circ} \mathrm{C}\) day, the interior of a house is maintained at \(22^{\circ} \mathrm{C}\) by the continuous operation of a 5 -ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs \(\$ 0.10\) per kWh and that the unit's coefficient of performance is 15\(\%\) that of an ideal refrigerator. \(1 \mathrm{kWh}=3.60 \times 10^{6} \mathrm{J}\)

(II) \(\mathrm{A} 2.8\) -kg piece of aluminum at \(43.0^{\circ} \mathrm{C}\) is placed in 1.0 \(\mathrm{kg}\) of water in a Styrofoam container at room temperature \(\left(20^{\circ} \mathrm{C}\right) .\) Estimate the net change in entropy of the system.

(III) The specific heat per mole of potassium at low temperatures is given by \(C_{V}=a T+b T^{3}, \quad\) where \(a=2.08 \mathrm{~mJ} / \mathrm{mol} \cdot \mathrm{K}^{2}\) and \(b=2.57 \mathrm{~mJ} / \mathrm{mol} \cdot \mathrm{K}^{4}\). Determine (by integration) the entropy change of \(0.15 \mathrm{~mol}\) of potassium when its temperature is lowered from \(3.0 \mathrm{~K}\) to \(1.0 \mathrm{~K}\).

(a) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are \(710^{\circ} \mathrm{C}\) and \(430^{\circ} \mathrm{C},\) and of the second \(415^{\circ} \mathrm{C}\) and \(270^{\circ} \mathrm{C}\). If the heat of combustion of coal is \(2.8 \times 10^{7} \mathrm{~J} / \mathrm{kg},\) at what rate must coal be burned if the plant is to put out \(950 \mathrm{MW}\) of power? Assume the efficiency of the engines is \(65 \%\) of the ideal (Carnot) efficiency. (b) Water is used to cool the power plant. If the water temperature is allowed to increase by no more than \(5.5 \mathrm{C}^{\circ}\), estimate how much water must pass through the plant per hour.

(II) Energy may be stored for use during peak demand by pumping water to a high reservoir when demand is low and then releasing it to drive turbines when needed. Suppose water is pumped to a lake 135 \(\mathrm{m}\) above the turbines at a rate of \(1.35 \times 10^{5} \mathrm{kg} / \mathrm{s}\) for 10.0 \(\mathrm{h}\) at night. (a) How much energy \((\mathrm{kWh})\) is needed to do this each night? \((b)\) If all this energy is released during a 14 -h day, at 75\(\%\) efficiency, what is the average power output?

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.