Question

An 7.5 g ice cube at −10°C is put into a Thermos flask containing 150 cm³ of water at 21°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

water is 333 kJ/kg.)

Answer 1

First we need to find the equilibrium temperature of the water, T

change in temperature: Q = m c DT

change in phase: Q = m L

Energy gained by ice

• Ice -10 °C to 0 °C: m = 7.5.0 g = 7.5´10^-3 kg, c = 2220 J.kg^-1.K^-1

Q₁ = m c DT = (7.5´10^-3)(2220)(10) = 166.5 J

• Ice to water: L = 3.33´10⁵ J.kg^-1

Q₂ = m L = (7.5´10^-3)(3.33´10⁵) = 2497.5 J

• Water 0 °C to T: m = 7.5´10^-3 kg, c = 4190 J.kg^-1.K^-1

Q₃ = m c ( T - 0 ) = (7.5´10^-3)(4190)T = 31.4T J

Energy lost by water

• Water from 21 °C to T: V = 150 cm³, m = 150 g = 0.150 kg, c = 4190 J.kg^-1.K^-1

Q₄ = m c ( 20 - T ) = (0.150)(4190)(21 - T) = 13198.5 - 628.5T J

Energy gained by ice = Energy lost by water

Q₁ + Q₂ + Q₃ = Q₄

166.5 + 2497.5 + (31.4 T) = 13198.5 - 628.5T J

659.9T = 10534.5

T = 15.953 °C = 12.2 + 273.15 = 288.96 K

Change in entropy - consider four reversible processes

T_i to T_f, DS = ò dQ/T = m c ò dT/T = m c ln( T_f / T_i )

(1) Ice -10 °C = 263 K to 0 °C = 273 K : m = 7.5´10^-3 kg, c = 2220 J.kg^-1.K^-1

DS₁ = (7.5´10^-3)(2220) ln( 273 / 263 ) = + 0.618 J.K^-1

(2) Ice to water 0 °C = 273 K: L = 3.33´105 J.kg^-1

DS₂ = Q / T = m L / T = (7.5´10^-3)(3.33´10⁵) / 273 = 9.14J.K^-1

(3) Water 0 °C = 273 to 285 K: m = 7.5´10^-3 kg, c = 4190 J.kg^-1.K^-1

DS₃ = (7.5´10^-3)(4190) ln( 285 / 273 ) = 1.35 J.K^-1

(4) Water from 21 °C = 294 K to 285 K: m = 0.150 kg, c = 4190 J.kg^-1.K^-1

DS₄ = (0.150)(4190) ln( 285 / 294 ) = - 19.54 J.K^-1

Change in entropy DS = DS₁ + DS₂ + DS₃ + DS₄

DS = + 0.618 + 9.14 + 1.35 - 19.54 = -8.43 J.K^-1