Question

If 2.50 m3 of water at 0∘C is frozen and cooled to -9.0 ∘C by being in contact with a great deal of ice at -9.0 ∘C, estimate the total change in entropy of the process.

Answer 1

For the water:

For phase change of water from liquid to solid,

ΔS_w1 = Q/T = -mL_f/T = (2.50 * 1000) * (334 * 10³) / 273 = -3.06 * 10⁶ J/K

For cooling of ice,

ΔS_w2 = ₂₇₃∫²⁶⁴ mC_icedT/T = (2.50 * 1000) * (2.03 * 10³) * ln(264 / 273) = -0.17 * 10⁶ J/K

So, total change in entropy for water, ΔS_w = -(3.06 + 0.17) * 10⁶ = -3.23 * 10⁶ J/K

For the great deal of ice:

ΔS_ice = m(L_f + C_iceΔT)/T = (2.50 * 1000) * [(334 * 10³) + (2.03 * 10³ * 9)] / 264 = 3.34 * 10⁶ J/K

So, total change in entropy of the system,

ΔS = ΔS_w + ΔS_ice = (-3.23 * 10⁶) + (3.34 * 10⁶) = 1.1 * 10⁵ J/K