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An 8.5 g ice cube at −10°C is put into a Thermos flask containing 150 cm3...

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Physics

An 8.5 g ice cube at −10°C is put into a Thermos flaskcontaining 150 cm3 of water at 30°C. By how much has theentropy of the cube-water system changed when equilibrium isreached? The specific heat of ice is 2220 J/kg · K. (The latentheat of fusion for ice is 333 kJ/kg and the specific heat of wateris 4187 J/kg · K.)

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4.2 Ratings (749 Votes)

The change in entropy in this case is


       ΔS1 = integral ( dQ /T)


              = m Cice ln (Tf / Ti)


              = (8.5 x 10-3) (2220) ln (273 / 263)


              = 0.704 J/ K


Now the ice melts at 0oC , Then


       ΔS2 = Q / T = mL /T   = 8.5 * 333 / 273 = 10.4 J /K


Now the water warms to lake temperature


         ΔS3 = m Cw ln(Tf / Ti)


                = (1000)(1.5*10-4 )(4187) ln (293 / 273)


                = 44.4 J /K


Therefore total change in entropy


         ΔS = 56.1 J /K
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