....o mobily 9:06 PM F flipitphysics.com Fliplt Physics Physics 4B: EM and Heat,

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....o mobily 9:06 PM F flipitphysics.com Fliplt Physics Physics 4B: EM and Heat, S2017 macmillan learning California State University, Fresno Homework: HW 6 of Ch 19: Entropy til Wednesday, March 59 PM Tipler6 19.P 060 A 78 g piece of ice at 0.0 C is placed in an insulated calorimeter of negligible heat capacity containing 100 g of water at 100 C. (a) What is the final temperature of the water once thermal equilibrium is established? C Submit You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question (b) Find the entropy change of the universe for this process. J/K Submi You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question. Copyright 017 Fre Worth Publish a division of Macmillan Learning About l Tech Support l Find Your Local Sales Rep I Terms of Use I Privacy Policy 39% Problems Print Assign Tipler 6 19.P.05 Standard Exercise Tipler6 19. P.06

Explanation / Answer

part a )

heat gain = heat lost

ice change from ice to water + ice water from 0 to t = water lost heat

m*Lf + m*c*t = m*c*(100-t)

78*10^-3 kg * 333.5 kJ/kg + 78 * 10^-3 * 4.18 kJ/kg*t = 100 x 10^-3 * (4.18 kJ/kg) (100-t)

26.013 + 0.32604t = 41.8 - 0.418t

0.74404t = 15.787

t = 21.2 oC

part b )

The entropy change oftheuniverse is the sum of the entropychanges of the ice and the water

dS_ice = phase change + to equib temp

dS_ice = mLf/Tf + m*c*ln(Tf/Ti)

dS_ice = 78x10^-3 *333.5/273 + 78*10^-3 * 4.18 * ln(294.2/273)

dS_ice = 119.67 J/K

dS_water = 0.100 kg * 4.18 * ln(294.2/373) = -99.2 J/K

dS_u = dS_ice + dS_water

dS_u = 20.47 J/K

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