Question

680 kg of fish at 5°C are to be frozen and stored at – 12°C. The specific heat of fish above freezing point is 3.182, and below freezing point is 1.717 kJ/kg K. The freezing point is – 2°C, and the latent heat of fusion is 234.5 kJ/kg. How much heat must be removed to cool the fish, and what per cent of this is latent heat?

Answer 1

Heat to be removed above freezing point 

= 680 × 3.182 × {5 – (-2)} kJ

= 15.146 MJ

Heat to be removed latent heat

= 680 × 234.5 kJ

= 159.460 MJ

Heat to be removed below freezing point

= 680 × 1.717 × {– 2 – (– 12)} kJ

= 11.676 MJ

∴ Total Heat = 186.2816 MJ

% of Latent heat = \(\frac{159.460}{186.2816}\) × 100

= 85.6 %