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

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

Expert Solution

The change in entropy in this case is

ΔS1 = integral ( dQ /T)

= m C_ice ln (Tf / Ti)

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

= 0.704 J/ K

Now the ice melts at 0^oC , 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

genius_generous answered 1 year ago

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