Freezing Technology

3.1.5 Principles of freezing

It is a common fact that the foods with high moisture content rapidly deteriorate in quality. During freezing, the water in the food is separated out from other food components, and is frozen. Thus, a food is protected, preserved from deteriorating influences such as temperature and water. The lower temperature slows down the rate of chemical reaction and water is also removed from the sphere of activity. The water is reabsorbed by the food components upon thawing and food is restored to its original quality. However, when a food is frozen, ie. When the water undergoes transition from liquid to solid state and also in subsequent thawing ie., when there is transition from solid to liquid state, there is a damaging effect on food. The texture and the structure of the food may be affected to such an extent that it may become irreversible and upon thawing the food will not be restored to its original condition.

                 Fresh fish flesh normally contains 60-80% water and the freezing process converts most of this water into ice. At normal atmospheric pressure, pure water will change from liquid to solid (ice) at 0^oC, i.e., it will freeze. The water in fish flesh contains salts and chemicals, which have the effect of lowering the temperature at which the water begins to freeze. The water in fish flesh begins to freeze at about –1^oC and as the temperature drops below –1^oC, more water is frozen out and the concentration of salts in the remaining water rises, so that its freezing point is lowered further. At –5^oC, it would appear that all the water is frozen, but over 20% of the water in fish muscle is still unfrozen. Even at –30^oC, approximately 10% of the water remains unfrozen.The above defined heat units are applied here.

                   In order to change the physical state of a substance from a liquid to a solid, energy or latent heat has to be removed from the substance. In order to lower the temperature of 1g of water by 1^oC, at temperatures above 0^oC, 1 calorie of heat must be removed this is known as the “specific heat”. However, to change water at 0^oC to ice at 0^oC, 80 calories must be removed for each g of water. In other words, the specific heat of liquid water is 1 and the latent heat of changing liquid water to ice is 80. The specific heat of ice at temperatures below 0^oC is 0.5, which means that to lower the temperature of ice by 1^oC, 0.5 calories of heat needs to be removed. For all practical purposes, it is assumed that fish have the same values for specific heat and latent heat as water. All this means that, if heat is removed from fish at a constant rate, there will be a period during the fish freezing, where the temperature of fish will not drop. This period lasts until approximately 75% of the water is frozen, when the temperature begins to drop again. There are three stages to freezing fish fig.1). During stage 1, the temperature falls fairly rapidly to just below 0^oC (and it is called as initial falling rate period), during stage 2, the temperature remains fairly constant at about –1^oC as the bulk of the water (three quarters of the water) in the fish freezes (this stage is known as the ‘thermal arrest’ period) and during stage 3 (final falling rate period), the temperature again drops and most of the remaining water becomes frozen. The temperature at –1.1^oC, freezing begins, and –5^oC, about 80% of water is frozen, this range is termed at ‘Critical range’. The ideal thermal arrest is only 30min. Freezing is complete only when the equilibrium temperature reaches –18^oC(ie. the process should not be regarded as complete unless and until the product temperature has reached –18^oC at the thermal centre after thermal stabilization). It also shows that even at temperatures as low at –30^oC, a portion of water in the fish muscle still remains in the unfrozen state. 

Under ideal conditions, fish should be frozen to –30^oC in 2 hours.              

 

Using simple mathematics, the amount of energy required to freeze fish can be calculated.

                 Suppose, if 1 kg fish at 25^oC is to be frozen at –30^oC. During stage 1, one calorie of energy has to be extracted for each gram material for each drop of 1^oC. In an example, the temperature of 1000g of fish will be lowered from 25^oC to –1^oC, ie., by 26^oC. The energy required will be equal to 1000 x 26 x the specific heat of water (1) = 26000 calories or 26 kilocalories (kcal).

During stage 2, 80 calories of energy needs to be extracted for each gram material frozen. In the example, 1000g of fish are to be frozen. The energy required will be equal to 1000 x the latent heat for freezing water (80), which equals 80,000 calories or 80kcal.

                 During stage3, 0.5 calories of energy needs to be extracted for each g material for each 1^oC drop in temperature. In the example, the temperature of 1000g of fish will be lowered from –1 to –30^oC, ie., by 29^oC. The energy required will be equal to 1000 x 29 x the specific heat of ice (0.5 = 14500 calories =14.5 kcal.

                 To summaries: 

                 Stage 1,       1000 x 26 x 1       =     26 kcal

                 Stage 2,       1000 x 80             =     80 kcal

                 Stage 3,       1000 x 29 x 0.5    =     14.5 kcal

                 Adding these three figures gives 120.5 kcal, ie., to freeze 1kg of fish from 25^oC to –30^oC, 120.5 kcal of energy is required.

                 From the above example, it is apparent that more than 50% of the energy extraction during freezing of fish is taking place in stage 2. The thermal arrest period, where little or no drop in temperature is occurring, and this period is a critical one, if a good frozen product is to be produced. Ideally, fish should pass through the thermal arrest period as quickly as possible, due to the following reasons:

1.   It is postulated that slow freezing produces large ice crystals in the cells of the fish, which can be larger than the cells themselves and so break the cell walls. But it is not so, because it is evident that the walls of fish muscle cells are sufficiently elastic to accommodate the larger ice crystals without excessive damage. Also, most of the water in fish muscle is bound to the protein in the form of a gel and little fluid would be lost.

2.      As the water begins to freeze in the flesh, the concentration of salts and chemicals in the remaining water rises. This high concentration of salts and enzymes can cause accelerated autolysis and denaturation of protein during slow freezing, This result in an inferior quality product and it is mainly due to denaturation of the protein. Changes take place in some fractions of the protein as a result of freezing and since the proteins are altered from their “native” state they may be said to be “denatured”, hence the term “protein denaturation”.

3.      At temperature around 0^oC, certain types of bacteria are still active and bacterial spoilage can still occur.

4.      Textural changes occur in fish, which have been frozen slowly, caused by the presence of large ice crystals and denaturation of protein during the accelerated autolysis. In addition, a phenomenon known as ‘thaw drip’ occurs, when slowly frozen fish are thawed. On thawing, the water which was originally bound within the cells are released and considerable loss in weight can occur.

                 However, from a textural point of view, it is unlikely that a highly trained taste panel could detect the difference between fish frozen in 1 hour and those frozen in 8 hours, Freezing times of 24 hours or more will almost certainly result in an inferior product and very long freezing times can result in bacterial spoilage making the fish unfit for consumption.