5.7: Questions
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- 20948
Deeper questions
The following questions require some thought and reaching the answer may require you to think beyond the contents of this TLP.
1. Crystallization is usually fatal to cells because:
3. Crystallization of minerals in the cells can be limited by:
In some alpine plants, extracellular ice formation occurs at around –2°C. What is the critical radius for ice nucleation at this temperature?
In these plants, the extracellular ice forms on ice nucleating agents. If an ice nucleating agent is a circular disc and is a perfect template for ice, what size must it be for nucleation to occur at –2°C?
(For the ice-water interface, the interfacial energy γ = 0.028 J m-2; the latent heat of freezing of ice is ΔHv = -3.34 x 108 J m-3.)
hint: The critical radius for nucleation is r*, where r* = -2γ /ΔGv, and γ is the solid-liquid interfacial energy and ΔGv is the free energy of freezing per unit volume. For further details see the section on nucleation and crystallization.
- Answer
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r*, the critical nucleus for nucleation is given by, r* = -2γ /ΔGv where γ is the solid-liquid interfacial energy and ΔGv is the free energy of freezing per unit volume. γ is given in the question and ΔGv can be determined from ΔHv. For freezing at a temperature T, the supercooling, ΔT is (Tm - T), where (Tm is the melting temperature. The entropy of fusion per unit volume is ΔSv = ΔHv/Tm. As shown in the section on Nucleation and Crystallization, ΔGv = ΔSvΔT. Thus:
\[r^{*}=-\frac{2 \gamma}{\Delta G_{v}}=-\frac{2 \gamma}{\Delta S_{v} \Delta T}=-\frac{2 \gamma T_{m}}{\Delta H_{v} \Delta T}\]
With γ and ΔHv as given, Tm taken to be 273 K, and T = 271 K, we find r* = 22.9 nm
For ice to grow on the INA, the radius of the INA must be greater than or equal to r*, so, for ice to grow, the INA should have radius 22.9 nm.