Subtopic: Preservation of food by cooling and freezing
Question Statement:
One ton of marine fishery products are to be brought down from 32 °C temperature to −18 °C in half an hour time using a plate freezer. The freezing point of 85 % water (based on total mass of the product) present in the fish is −0.3 °C. Specific heat capacities of fresh and frozen fish solids (15 % of total mass) are 3.2 and 1.8 kJ kg⁻¹ K⁻¹, respectively. Specific heat capacity of fresh water is 4.2 kJ kg⁻¹ K⁻¹ and that of ice is 2.2 kJ kg⁻¹ K⁻¹. Latent heat of crystallization of water is 335 kJ kg⁻¹. For freezing the products, the compressor power consumption with a vapour compression refrigeration cycle (coefficient of performance, 3.66) is _________ kW.
Official Answer Key:
68.00 to 69.00
Step-by-Step Solution & Explanation
Mass split: water = 0.85 × 1000 = 850 kg; solids = 150 kg.
Stage 1 — sensible cooling above freezing (32 → −0.3 °C, ΔT = 32.3 K):
Q₁ = 850 × 4.2 × 32.3 + 150 × 3.2 × 32.3 = 115 318 + 15 504 = 130 822 kJ
Stage 2 — latent heat of crystallization:
Q₂ = 850 × 335 = 284 750 kJ
Stage 3 — sensible cooling below freezing (−0.3 → −18 °C, ΔT = 17.7 K):
Q₃ = 850 × 2.2 × 17.7 + 150 × 1.8 × 17.7 = 33 099 + 4 779 = 37 878 kJ
Total Q = 130 822 + 284 750 + 37 878 = 453 450 kJ in 1800 s.
Refrigeration load = 453 450/1800 = 251.9 kW.
Compressor power = Load/COP = 251.9/3.66 = 68.8 kW — within the official range 68.00–69.00.
Common error: omitting Stage 3 or applying the frozen-solids c_p to the whole mass.
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