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Chapter 4: Problem 132

What are the factors that affect the quality of frozen fish?

Short Answer

Expert verified
Answer: Some key factors that influence the quality of frozen fish include freezing rate and temperature, storage temperature, time of storage, handling and packaging, and the species of fish and their initial quality. A rapid freezing rate at low temperatures helps preserve texture and quality by minimizing ice crystal growth. Maintaining a consistent and low storage temperature slows down processes that can cause spoilage. Consuming fish products within their recommended shelf-life helps maintain optimal quality. Proper handling and packaging minimize temperature fluctuations, contamination, and damage during transportation, storage, and display. The fish species and their initial quality—ensuring proper handling from harvest to freezing—also play a role in the final quality of frozen fish products.

Step by step solution

01

Identify the factors that affect the quality of frozen fish

To understand the factors that affect the quality of frozen fish, it is important to know the potential sources of quality degradation. Some of the factors that are involved in the quality of frozen fish include: 1. Freezing rate and temperature 2. Storage temperature 3. Time of storage 4. Handling and packaging 5. Species of fish and their initial quality
02

Explain the impact of freezing rate and temperature

The freezing rate and temperature are crucial factors that can affect the frozen fish quality. A rapid freezing rate at low temperatures helps to minimize the growth of ice crystals within the fish tissue, which can cause damage to the cell structure. Smaller ice crystals preserve the texture and quality of the fish.
03

Explain the significance of storage temperature

Maintaining a consistent and low storage temperature is essential for preserving the quality of frozen fish. Lower storage temperatures help to slow down the biochemical and microbiological processes that can cause spoilage and quality deterioration. A temperature of -18 degrees Celsius or lower is generally considered appropriate for the long-term storage of frozen fish products.
04

Discuss the role of storage time

Over time, even at low temperatures, frozen fish can undergo a gradual decline in quality. Factors like enzyme activity, oxidation, and moisture migration may continue to cause spoilage and texture changes over time. Therefore, it is important to ensure that fish products are consumed within their recommended shelf-life to maintain optimal quality.
05

Explain the importance of handling and packaging

Proper handling and packaging can play a vital role in maintaining the quality of frozen fish. Minimizing temperature fluctuations, using appropriate packaging materials, and preventing contamination from foreign substances help to preserve the quality and safety of the fish during transportation, storage, and retail display.
06

Take into account species of fish and their initial quality

The quality of frozen fish products can be influenced by the species of fish and their initial quality prior to freezing. Some fish species are more susceptible to quality changes during freezing and storage than others. It is also crucial to ensure that the fish are handled properly from harvest to freezing to maintain their quality. This includes factors such as proper gutting, washing, chilling, and inspecting for defects or damages.

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Most popular questions from this chapter

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

The chilling room of a meat plant is \(15 \mathrm{~m} \times 18 \mathrm{~m} \times\) \(5.5 \mathrm{~m}\) in size and has a capacity of 350 beef carcasses. The power consumed by the fans and the lights in the chilling room are 22 and \(2 \mathrm{~kW}\), respectively, and the room gains heat through its envelope at a rate of \(14 \mathrm{~kW}\). The average mass of beef carcasses is \(220 \mathrm{~kg}\). The carcasses enter the chilling room at \(35^{\circ} \mathrm{C}\), after they are washed to facilitate evaporative cooling, and are cooled to \(16^{\circ} \mathrm{C}\) in \(12 \mathrm{~h}\). The air enters the chilling room at \(-2.2^{\circ} \mathrm{C}\) and leaves at \(0.5^{\circ} \mathrm{C}\). Determine \((a)\) the refrigeration load of the chilling room and \((b)\) the volume flow rate of air. The average specific heats of beef carcasses and air are \(3.14\) and \(1.0 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the density of air can be taken to be \(1.28 \mathrm{~kg} / \mathrm{m}^{3}\).

What is the effect of cooking on the microorganisms in foods? Why is it important that the internal temperature of a roast in an oven be raised above \(70^{\circ} \mathrm{C}\) ?

A thermocouple, with a spherical junction diameter of \(0.5 \mathrm{~mm}\), is used for measuring the temperature of hot air flow in a circular duct. The convection heat transfer coefficient of the air flow can be related with the diameter \((D)\) of the duct and the average air flow velocity \((V)\) as \(h=2.2(V / D)^{0.5}\), where \(D, h\), and \(V\) are in \(\mathrm{m}, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{m} / \mathrm{s}\), respectively. The properties of the thermocouple junction are \(k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8500 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Determine the minimum air flow velocity that the thermocouple can be used, if the maximum response time of the thermocouple to register 99 percent of the initial temperature difference is \(5 \mathrm{~s}\).

Copper balls \(\left(\rho=8933 \mathrm{~kg} / \mathrm{m}^{3}, k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=\right.\) \(\left.385 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, \alpha=1.166 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at \(200^{\circ} \mathrm{C}\) are allowed to cool in air at \(30^{\circ} \mathrm{C}\) for a period of 2 minutes. If the balls have a diameter of \(2 \mathrm{~cm}\) and the heat transfer coefficient is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the center temperature of the balls at the end of cooling is (a) \(104^{\circ} \mathrm{C}\) (b) \(87^{\circ} \mathrm{C}\) (c) \(198^{\circ} \mathrm{C}\) (d) \(126^{\circ} \mathrm{C}\) (e) \(152^{\circ} \mathrm{C}\)
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