EXERCISECompare the cost to make 48 kg of ice using two different refrigerators; one has a COP of 5.00 and the other has a COP of6.00. Electricity costs $0.10 per kWh. Using the same assumptions given in the example, find the cost difference (indollars) between the refrigerators to freeze the water.Note: A kilowatt-hour [kWh] is a unit of energy. It is the amount of energy an appliance rated at 1 kW (1000 W) wouldconsume if it ran for 1 hour. Therefore, 1 kWh = 3.60x106 J, as evident in the following unit conversion:1 kWh x1000W J/s 3600sh(hour)XkWWX= 3,600,000 JHint: The cost difference will be a very small amount. Enter your answer using three significant figures.Hintcost difference = $ Freezing WaterA certain refrigerator has a COP of 5.00. When the refrigerator is running, its power input is 500 W. A sample of water ofmass 700 and temperature 20.0°C is placed in the freezer compartment. How long does it take to freeze the water to iceat 0°C? Assume all other parts of the refrigerator stay at the same temperature and there is no leakage of energy from theexterior, so the operation of the refrigerator results only in energy being extracted from the water.SOLUTIONConceptualize Energy leaves the water, reducingits temperature and then freezing it into ice. The time intervalrequired for this entire process is related to the rate at which energy is withdrawn from the water, which, in turn, is relatedto the power input of the refrigerator.Categorize We categorize this example as one that combines our understanding of temperature changes and phasechanges from Chapter 20 and our understanding of heat pumps from this chapter.AnalyzeUse the power rating of the refrigerator to find the time interval required for the freezing process to occur. (Use thefollowing as necessary: P and W.)WP =AtThe COP for a freezer isΔt =→ At =COP = 1QcI/W = 1Qcl/ (P At)Use Q = mcAT and Q = LAm to substitute the amount of energy IQ that must be extracted from the water of mass m.(Use the following as necessary: L₁, Am, and AT.)At =WPmcAT +At =LAm✓P(COP)Recognize that the amount of water that freezes is Am = -m because all the water freezes. (Use the following asnecessary: L₁, m, c and AT.)- m( | CAT+L₁ | x )|ΔΤP(COP)Substitute numerical values to find the time interval. (Enter your answer in seconds. Assume the specific heat of water is4,186 J/(kg . °C), and the latent heat of fusion of water is 3.33 x 105 J/kg.)✓SFinalize In reality, the time interval for the water to freeze in a refrigerator is much longer than this, which suggests thatthe assumptions of our model are not valid. Only a small part of the energy extracted from the refrigerator interior in agiven time interval comes from the water. What other sources of energy must be extracted by the refrigerator? (Select allthat apply.)O Energy must also be extracted from the container in which the water is placed.O Energy from the environment must be extracted into the refrigerator.O Energy that continuously leaks into the interior must be extracted.EXERCISECompare the cost to make 48 kg of ice using two different refrigerators; one has a COP of 5.00 and the other has a COP of6.00. Electricity costs $0.10 per kWh. Using the same assumptions given in the example, find the cost difference (indollars) between the refrigerators to freeze the water.Note: A kilowatt-hour [kWh] is a unit of energy. It is the amount of energy an appliance rated at 1 kW (1000 W) would

Given data: Mass of ice, m = 48kg COP1 = 5 COP2 = 6 Electricity cost = 0.1$ / kWh

Compare the cost to make 48 kg of ice using two different refrigerators; one has a COP of 5.00 and the other has a COP of 6.00. Electricity costs $0.10 per kWh. Using the same assumptions given in the example, find the cost difference (in dollars) between the refrigerators to freeze the water. Note: A kilowatt-hour [kWh] is a unit of energy. It is the amount of energy an appliance rated at 1 kW (1000 W) would consume if it ran for 1 hour. Therefore, 1 kWh = 3.60x106 J, as evident in the following unit conversion: 1 kWh x 1000W J/s 3600s X X h(hour) KW W = 3,600,000 J Hint: The cost difference will be a very small amount. Enter your answer using three significant figures. Hint cost difference = $

Compare the cost to make 48 kg of ice using two different refrigerators; one has a COP of 5.00 and the other has a COP of 6.00. Electricity costs $0.10 per kWh. Using the same assumptions given in the example, find the cost difference (in dollars) between the refrigerators to freeze the water. Note: A kilowatt-hour [kWh] is a unit of energy. It is the amount of energy an appliance rated at 1 kW (1000 W) would consume if it ran for 1 hour. Therefore, 1 kWh = 3.60x106 J, as evident in the following unit conversion: 1 kWh x 1000W J/s 3600s X X h(hour) KW W = 3,600,000 J Hint: The cost difference will be a very small amount. Enter your answer using three significant figures. Hint cost difference = $

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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