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WileyPLUs f Li edugen.wileyplus.com/edugen/student/mainfr.un. M Inbox rhino291 yanortizsf Yah Ryan Ortiz Out Canvas WileyPlus . Google Drive M Inbox du cathol a Amazon.com PioneerWeb Un e Chegg Facebook y Twitter OrgSync × Chegg Study l Guide x Other bookmarks WileyPLUS: MyWileyPLUS| Help Contact Us I Log Out WileyPLUS aliday, Fundamentals of Physics, 10e Halliday, Fundamentals of Physics, 10e University Physics (PHYS 1211-1214) Home Read, Study& Practice Assignment Gradebook ORION Assignment > Open Assignment FULL SCREEN PRINTER VERSION BACK NEXT ASSIGNMENT RESOURCES Chapter 18, Problem 035 Exam 3 Review (Ch18 - ermO Your answer is partially correct. Try again. le An insulated Thermos contains 160 cm3 of hot coffee at 84.0°C. You put in a 13.0 g ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment. The specific heat of water is 4186 J/kg K. The latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm3 le Number 711.62 Unitsco na le SHOW HINT Chapter 18, Problem 035 le LINK TO TEXT LINK TO SAMPLE PROBLEM le Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER Review Score Copyright © 2000-2016 by John Wiley & Sons, Inc. or related companies. All rights reserved Review Results by Study Obiective 5:08 PM 5/24/2016

Explanation / Answer

Inital Temp of Coffee, = 84.0°C
Mass of ice, mi = 13 g
Latent Heat of ice, Li = 333.0 J/gm
Specific Heat of Water, Cw = 4.186 J/gm k

Calculating Mass of coffee,
Density = Mass/Volume
1 g/cm^3 = m /160 cm^3
m = 160 g

Now Let the thermal equilibrium temperature of the mixture be Tf,
Heat Lost by Ice, = Heat gained by Coffee
mi * Li + mi*Cw*Tf = m * Cw * (84 - Tf)
13 * 333 + 13 * 4.186 * Tf = 160 * 4.186 * (84 - Tf)
Tf = 71.7 °C

Now they are asking, by how many degree coffee is cooled,

So Answer is , 84.0 - 71.7 = 12.3 °C

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