A food material having initial moisture content of 400% (dry weight basis) is poured into a tray of freeze dryer which operates at 40 Pa in 10 mm layers. It is to be dried to 8% moisture (dry weight basis) at a maximum surface temperature of 55 °C. The dried food has a thermal conductivity of 0.03 W m⁻¹ K⁻¹, a density of 470 kg m⁻³, a permeability of 2.4×10⁻⁸ kg s⁻¹ m⁻¹, and latent heat of sublimation of 2.95×10³ kJ kg⁻¹. It is assumed that the pressure at the ice front remains constant at 78 Pa. The temperature at the sublimation front will be
Options:
(A) −73.5 °C
(B) −35.7 °C
(C) −25.28 °C
(D) −15.72 °C
Official Answer Key:
C
Step-by-Step Solution & Explanation
The sublimation (ice) front is at the saturation vapour pressure of ice corresponding to 78 Pa. Using the ice sublimation vapour-pressure–temperature relation, a saturation pressure of 78 Pa corresponds to a temperature of about −25 °C, matching option (C). (An independent check using the approximate Clausius–Clapeyron form for ice gives a value in the same range, −23 to −25 °C, consistent with the key.)
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