/*! 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 40 A newly proposed device for gene... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A newly proposed device for generating electricity from the sun is a heat engine in which the hot reservoir is created by focusing sunlight on a small spot on one side of the engine. The cold reservoir is ambient air at \(20^{\circ} \mathrm{C}\). The designer claims that the efficiency will be \(60 \% .\) What minimum hot-reservoir temperature, in \({ }^{\circ} \mathrm{C},\) would be required to produce this efficiency?

Short Answer

Expert verified
459.73 °C

Step by step solution

01

Convert Cold Reservoir Temperature to Kelvin

The temperature of the cold reservoir is given in Celsius. For the formula we need it in Kelvin. The direct conversion from Celsius to Kelvin is obtained by adding 273.15 to the Celsius temperature. Therefore, the cold reservoir temperature in Kelvin is \(20 + 273.15 = 293.15\) K.
02

Apply the Efficiency Formula to Find the Hot Reservoir Temperature

By rearranging the equation \( \eta = 1 - \frac{T_c}{T_h} \) for \(T_h\), we get \(T_h = \frac{T_c}{1-\eta}\). We know the efficiency η is 60%, or 0.60 in decimal form. Substituting in the values \(T_c = 293.15\) K and \(η = 0.60\), we get \(T_h = \frac{293.15}{1-0.60} = 732.88\) K.
03

Convert Hot Reservoir Temperature to Celsius

We have found the hot reservoir temperature in Kelvin, but the question requests the answer in Celsius. So we need to convert the Kelvin temperature to Celsius by subtracting 273.15 from it. Thus, the temperature in Celsius of the hot reservoir required to maintain the claimed efficiency is \(732.88 - 273.15 = 459.73^{\circ} \mathrm{C}\).

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 the relationships between heat, work, temperature, and energy. The laws of thermodynamics are fundamental rules that all natural processes follow. In the context of a heat engine, such as that described in the exercise, thermodynamics tells us that energy cannot be created or destroyed, only converted from one form to another.

Heat engines work by transferring energy from a hot reservoir (usually a source of heat) to a cold reservoir (typically a cooler environment) and in the process converting some of this heat energy into mechanical work. The performance of a heat engine is quantified by its efficiency, which is a measure of how well an engine converts heat into work. The Second Law of Thermodynamics implies that it is impossible for a heat engine to convert all of the thermal energy into work, and thus no heat engine can be 100% efficient. The exercise illustrates the implications of thermodynamics for engineering a practical device to harvest energy from the sun.
Temperature Conversion
Understanding how to convert temperature from one unit to another is crucial in the field of thermodynamics, especially when dealing with heat engines. The Kelvin scale is an absolute temperature scale used widely in the sciences. It starts at absolute zero, the theoretical point where particles have minimum thermal motion.

In temperature conversion, degrees Celsius can be converted to Kelvin and vice versa. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature, and to convert Kelvin back to Celsius, you subtract 273.15 from the Kelvin temperature. This step is often necessary in thermodynamic calculations since many formulas, like those for efficiency, require temperature to be in Kelvin. For the exercise, accurately converting the temperatures was necessary to assess the minimum temperature of the hot reservoir needed for the proposed efficiency.
Energy Conversion Efficiency
Energy conversion efficiency is a dimensionless number between 0 and 1 or, when expressed as a percentage, between 0% and 100%. It represents the fraction of energy that is converted to a desirable form when compared to the total energy input into the system. In the case of heat engines, the efficiency (\( u \) is calculated using the formula \( u = 1 - \frac{T_c}{T_h} \) where \( T_c \) and \( T_h \) are the cold and hot reservoir temperatures, respectively, in Kelvin.

The higher the efficiency, the better the system is at converting heat to work. However, according to the Carnot efficiency, the maximum possible efficiency of a heat engine depends on the temperatures of the hot and cold reservoirs and can never reach 100%. Real-world efficiencies are often much lower, as engines experience various losses. The claimed 60% efficiency of the device in the exercise implies it is quite efficient, suggesting a significant difference between the temperatures of the hot and cold reservoirs to achieve such performance.

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

\(10 \mathrm{J}\) of heat are removed from a gas sample while it is being compressed by a piston that does \(20 \mathrm{J}\) of work. What is the change in the thermal energy of the gas? Does the temperature of the gas increase or decrease?

A sleeping \(68 \mathrm{kg}\) man has a metabolic power of \(71 \mathrm{W}\). How many Calories does he burn during an 8.0 hour sleep?

An engine operating at maximum theoretical efficiency whose cold-reservoir temperature is \(7^{\circ} \mathrm{C}\) is \(40 \%\) efficient. By how much should the temperature of the hot reservoir be increased to raise the efficiency to \(60 \% ?\)

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story. If a \(1500 \mathrm{kg}\) car is driven in sand at \(5.0 \mathrm{m} / \mathrm{s},\) the coefficient of rolling friction is \(0.06 .\) In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem. a. What propulsion force is needed to keep the car moving forward at a constant speed? b. What power is required for propulsion at \(5.0 \mathrm{m} / \mathrm{s} ?\) c. If the car gets 15 mpg when driving on sand, what is the car's efficiency? One gallon of gasoline contains \(1.4 \times 10^{8} \mathrm{J}\) of chemical energy.

Each time he does one pushup, Jose, who has a mass of \(75 \mathrm{kg}\), raises his center of mass by \(25 \mathrm{cm} .\) He completes an impressive 150 pushups in 5 minutes, exercising at a steady rate. a. If we assume that lowering his body has no energetic cost, what is his metabolic power during this workout? b. In fact, it costs Jose a certain amount of energy to lower his body \(-\) about half of what it costs to raise it. If you include this in your calculation, what is his metabolic power?

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.