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# A 4.50-kg block of ice al 0.00°C falls into the ocean and ISBN: 9780321675460 31

## Solution for problem 22E Chapter 20

University Physics | 13th Edition

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Problem 22E

A 4.50-kg block of ice al 0.00°C falls into the ocean and melts. The average temperature of the ocean is 3.50°C, including all the deep water. By how much does the melting of this ice change the entropy of the world? Does it make it larger or smaller? (?Hint?: Do you think that the ocean will change temperature appreciably as the ice melts?)

Step-by-Step Solution:

Solution 22E Step 1 of 6: As the ocean is very large compared to 4.5kg ice block, the immense ocean does not change its temperature. But ocean losses some entropy as it gives out some heat to melt 0 the ice and also to warm it up to 3.5 c. Whereas the ice gains entropy by absorbing heat 0 0 from the ocean and also as it warms from 0 c to 3.5 c. In order to find the total entropy change of the world, first we need to calculate entropy change in ice and also in ocean. Step 2 of 6: Given data, For ice, 0 Temperature, T ice 0 c T ice= 0 + 273.15 = 273.15 K Mass of ice, m= 4.5 kg 3 Latent heat of fusion, L=334× 10 J/kg Specific heat of water, C= 4190 J/kg.K For ocean, 0 Temperature, T ocean 3.5 c Tocean= 3.5 + 273.15 = 276.65 K Change in temperature, T = T ocean Tice= 276.65 K 273.15 K T = 3.5 K To find, Heat gained by ice during melting, Q melt= Entropy change in ice during melting, S melt Heat gained by ice while temperature is increasing, Q = temp Entropy change in ice while temperature is increasing, Stemp= Heat lost by ocean, Qocean= Q melt+ Qtemp= Entropy change in case of ocean, S = ocean Total entropy change in world, S world S melt+ S temp+ Socean= Step 3 of 6: To calculate heat gained and entropy change during melting of ice, The heat gain during the melting of 4.5kg ice block is, Qmelt= mL 3 Substituting m= 4.5 kg and L=334× 10 J/kg Q = (4.5 kg)(334 × 10 J/kg) melt 6 Q melt= 1.503 × 10 J Change in entropy during melting of ice is given by, S = Qmelt melt T ice 6 Substituting Q melt 1.503 × 10 J and T ice= 273.15 K S = 1.503×10 J melt 273.15 K 3 S melt 5.502 × 10 J/K

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##### ISBN: 9780321675460

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