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Chrome File Edit View History Bookmarks People Window Help 00% Mon 18 AM a E Chegg Study I Guided solution x Wiley PLUS C secure https edugen.wileyplus.co m/edugen/student/mainfrun gnment FULL SCREEN PRINTER VERSION BACK NEXT Chapter 20, Problem 009 Your answer is partially correct Try again. A 56 g ice cube at -82 C is placed in a lake whose temperature is 39 C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg K. (Hint: Will the ice cube affect the temperature of the lake?) Numbe 9.33 unit UK a the tolerance is +/-2% SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI LECTUR Question Attempts: 1 of 6 used SAVE FOR LATER SUBMIT ANSWER Show All Flash Player (16) dmg

Explanation / Answer

lake has way more of water then the mass of ice.

hence change in tenmp of lake will negligble.

so final temp of ice will be 39 deg C .

from -82 deg C to 0 deg C .

deltaS1 = m C ln(Tf/Ti)

= 0.056 x 2220 x ln[(273+0)/(273-82]

= 44.4 J/K

for state change at 0 deg C,

deltaS2 = Q / T = m Lf / T = (56 x 333.55) / (273 + 0) = 68.4 J/K

for raise in temp to 39 deg C

deltaS3 = 0.056 x 4186 x ln[(273 + 39)/273] = 31.30 J/K

for lake:

energy released = 0.056(2260 x 82 + 333550 + 4186 x 39 )

= 38198.9 J

deltaS4 = - 38198.9 / (273 + 39) = - 122.4 J/K

deltaS = S1 + S2 + S3 + S4 = 21.7 J/K

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