Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Question
Chapter 20, Problem 38P
To determine
To Show:The cost of air conditioning isproportional to
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
A large electrical power station generates 1000 MW of electricity with an efficiency of 35.0%. (a) Calculate the heat transfer to the power station, Qh, in one day. (b) How much heat transfer Qc occurs to the environment in one day? (c) If the heat transfer in the cooling towers is from 35.0C water into the local air mass, which increases in temperature from 18.0C to 20.0C, what is the total increase in entropy due to this heat transfer? (d) How much energy becomes unavailable to do work because of this increase in entropy, assuming an 18.0C lowest temperature? (Part of Qc could be utilized to operate heat engines or for simply heating the surroundings, but it rarely is.)
A typical coal-fired power plant generates 1000 MW of usablepower at an overall thermal efficiency of 40%. (a) What is the rateof heat input to the plant? (b) The plant burns anthracite coal, which hasa heat of combustion of 2.65 * 10^7 J/kg. How much coal does the plantuse per day, if it operates continuously? (c) At what rate is heat ejectedinto the cool reservoir, which is the nearby river? (d) The river is at18.0C before it reaches the power plant and 18.5C after it has receivedthe plant’s waste heat. Calculate the river’s flow rate, in cubic meters persecond. (e) By how much does the river’s entropy increase each second?
A large electrical power station generates 1000 MW of electricity with an efficiency of 35.0%. (a) Calculate the heat transfer to the power station, Qh , in one day. (b) How much heat transfer Qc occurs to the environment in one day? (c) If the heat transfer in the cooling towers is from 35.0º C water into the local air mass, which increases in temperature from 18.0º C to 20.0º C , what is the total increase in entropy due to this heat transfer? (d) How much energy becomes unavailable to do work because of this increase in entropy, assuming an 18.0º C lowest temperature? (Part of Qccould be utilized to operate heat engines or for simply heating the surroundings, but it rarely is.)
Chapter 20 Solutions
Physics for Scientists and Engineers
Ch. 20 - Prob. 1PCh. 20 - Prob. 2PCh. 20 - Prob. 3PCh. 20 - Prob. 4PCh. 20 - Prob. 5PCh. 20 - Prob. 6PCh. 20 - Prob. 7PCh. 20 - Prob. 8PCh. 20 - Prob. 9PCh. 20 - Prob. 10P
Ch. 20 - Prob. 11PCh. 20 - Prob. 12PCh. 20 - Prob. 13PCh. 20 - Prob. 14PCh. 20 - Prob. 15PCh. 20 - Prob. 16PCh. 20 - Prob. 17PCh. 20 - Prob. 18PCh. 20 - Prob. 19PCh. 20 - Prob. 20PCh. 20 - Prob. 21PCh. 20 - Prob. 22PCh. 20 - Prob. 23PCh. 20 - Prob. 24PCh. 20 - Prob. 25PCh. 20 - Prob. 26PCh. 20 - Prob. 27PCh. 20 - Prob. 28PCh. 20 - Prob. 29PCh. 20 - Prob. 30PCh. 20 - Prob. 31PCh. 20 - Prob. 32PCh. 20 - Prob. 33PCh. 20 - Prob. 34PCh. 20 - Prob. 35PCh. 20 - Prob. 36PCh. 20 - Prob. 37PCh. 20 - Prob. 38PCh. 20 - Prob. 39PCh. 20 - Prob. 40PCh. 20 - Prob. 41PCh. 20 - Prob. 42PCh. 20 - Prob. 43PCh. 20 - Prob. 44PCh. 20 - Prob. 45PCh. 20 - Prob. 46PCh. 20 - Prob. 47PCh. 20 - Prob. 48PCh. 20 - Prob. 49PCh. 20 - Prob. 50PCh. 20 - Prob. 51PCh. 20 - Prob. 52PCh. 20 - Prob. 53PCh. 20 - Prob. 54PCh. 20 - Prob. 55PCh. 20 - Prob. 56PCh. 20 - Prob. 57PCh. 20 - Prob. 59P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- Suppose you want to operate an ideal refrigerator with a cold temperature of 10.0C, and you would like it to have a coefficient of performance 7.00. What is the hot reservoir temperature for such a refrigerator?arrow_forwardCheck your Understanding Show that QhQh=QcQc for the hypothetical engine of Figure 4.10 The second property to be demonstrated is that all reversible engines operating between the same two reservoirs have the same efficiency. To this, stat with the two engines D and E of Figure 4.10 (a), which are operating between two common heat reservoirs at temperatures Th and Tc . First, assume that D is a reversible engine and that E is a hypothetical irreversible engine that has a higher efficiency than D. If both engines perform the same amount of work W per cycle, it follows from Equation 4.2 that QhQh . It then follows from the first law that QcQc . Figure 4.10 (a) Two uncoupled engines D and E working between the same reservoirs. (b) The engines, With D working reverse. Suppose the cycle of D is so that it operates as a refrigerator, and the two engines are coupled such that the work output of E is used to drive D, as shown in Figure 4.10(b). Since QhQh and QcQc , the net result of each cycle is equivalent to a spontaneous transfer of heat from the cold reservoir to the hot reservoir, a process second law does not allow. The original assumption must therefore be wrong, and it is impossible to construct an irreversible engine such that E is more efficient than the reversible engine D. Now it is quite easy to demonstrate that the efficiencies of all reversible engines operating between the same reservoirs are equal. Suppose that D and E are reversible engines. If they are as shown in Figure 4.10(b), the efficiency of E cannot be greater than the efficiency of D, or second law would violated. If both engines are then reversed, the same reasoning implies that the efficiency of D cannot be greater than the efficiency of E. Combining these results leads to the conclusion that all reversible engines working between same two reservoirs have the same efficiency.arrow_forwardIn a quasi-static isobaric expansion, 500 J of work are done by the gas. The gas pressure is 0.80 atm and it was originally at 20.0 L. If the internal energy of the gas increased by 80 J in the expansion, how much heat does the gas absorb?arrow_forward
- The temperature of the cold and hot reservoirs between which a Carnot refrigerator operates are 73 and 270 , respectively. Which is its coefficient of performance?arrow_forwardCalculate the net work output of a heat engine following path ABCDA as shown below.arrow_forwardCheck Your Understanding A Carnot refrigerator operates between two heat reservoirs whose temperatures 0 and 25 . (a) What is the coefficient of performance of the refrigerator? (b) If 200 J of work done on the working substance cycle, how heat per cycle is from cold reservoir? (c) How much heat cycle is discarded to the hot reservoir?arrow_forward
- The energy output of a heat pump is greater than the energy used to operate the pump. Why doesn't this statement violate the first law of thermodynamics?arrow_forwardCheck Your Understanding What is the efficiency of a perfect heat engine? What is the coefficient of performance of a perfect refrigerator?arrow_forwardA Carnot refrigerator, working between 0 and 30 is used to cool a bucket of water containing 102 m3 of water at 30 to 5 in 2 hours. Find the total amount of work needed.arrow_forward
- An engineer must design a refrigerator that does 300 J of work per cycle to extract 2100 J of heat per cycle from a freezer whose temperature is 10 . What is the maximum air temperature for which this condition can be met? Is this a reasonable condition to impose on the design?arrow_forwardIs it possible for the efficiency of a reversible engine to greater than 1.0? Is it possible for the coefficient of performance of a reversible refrigerator to be less than 1.0?arrow_forwardA Carnot heat pump operates between 0 and 20 . How much heat is exhausted into the interior of a house for every 1.0 J of work done by the pump?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
The Second Law of Thermodynamics: Heat Flow, Entropy, and Microstates; Author: Professor Dave Explains;https://www.youtube.com/watch?v=MrwW4w2nAMc;License: Standard YouTube License, CC-BY