Question

680 kg of fish at 5°C are to be frozen and stored
at - 12°C. The specific heat of fish above freez-
ing point is 3.182, and below freezing point is
1.717 kJ/kgK. 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?
Ans. 186.28 MJ, 85.6%​

Answer 1

Answer:

186.28 MJ, 85.6%​

Explanation:

Formula for specific heat is;

C= Q/ (m* delta T)

Heat to be removed above freezing point  

= 680 × 3.182 × {5 – (-2)} kJ =  

15.146 MJ  

Formla for latent heat is ;

L = Q/m ;

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 = 11.676 +159.46+15.146 = 186.2816 MJ

% of Latent heat  =159.460/186.28 x 100

                                =  85.6 %