Lab 3 Freezing and Thawing (1)

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University of Saskatchewan *

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345

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Chemistry

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Feb 20, 2024

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FABS 345.3 Unit Operations in Food Processing Lab #3 Freezing and Thawing of Food Products Meet in room 3E75 . Objectives: To familiarize students with aspects of the freezing process, to demonstrate the effects of freezing rate on product quality, and to demonstrate principles of freezing point depression, freeze concentration and supercooling. Week One : (Students working in groups) Part A. Freezing and Thawing 1. Each group will obtain approximately 220 grams of each of strawberries, and mushrooms in preparation for freezing. Dip in a bowl of water to clean them and remove from the water promptly. Let them drain in a colander or on some paper towels (don’t let them soak too long). 2. Remove the stem and leaves of the strawberry. Cut strawberry and mushroom in half (from center). 3. Obtain a Ziploc bag with two pieces of folded paper towel in it as packing material, and weigh the package weight (W pack ). Load ~100 g sample in the bag on the paper towel with fruit’s cut surface down and weigh the bag again (W pack+sample ). Prepare four bags for the two samples. 4. You will use two packages of each product for: a) freezing at a slow rate (inside an insulated container in a -18˚C chest freezer) and b) freezing at a medium rate (-30˚C in a blast freezer). 5. Label all bags of product with group name, weight, date and treatment. 6. Put them in separate tubs for the different freezing rate treatments, then the TA will put them in freezer. (We’ll bring them out next week for Part 2) 1

Part B. Freezing potato in sodium chloride brine solution 1. Each group will be given a big potato. On a cutting board, cut the two sides to form a 5-6 cm thickness slab. Using a hole puncher, cut out two pieces of potato cores (5 – 6 cm long, ~ 2 cm diameter). 2. Accurately measure the diameter of the potato cores. 3. Each group will obtain an insulated container with cold sodium brine solution in it (~ -15 o C). Measure and record the initial temperature of the brine solution. 4. Place the thermocouple-probe in the center of the potato core, inserting the probe to the half way point of the cylinder’s height. You can use a ruler to measure the height of the core, and your finger on the probe to mark the half-height during insertion. Put the potato core in the brine solution. 5. Record the temperature every 30 seconds until the temperature dropped to 5 o C; every 15 seconds until the temperature dropped to 0 o C; every 5 seconds until the temperature dropped to - 2 o C; every 15 seconds until the temperature dropped to - 5 o C; and then stop recording. 6. Measure and record the final temperature of the brine solution. Part C. Freezing curve 1. The lab instructor will prepare one can of each single strength iced tea, and concentrated iced tea. A thermocouple will be placed in the can. 2. Freeze at slow and medium rates as in part A (freezing data will be recorded and provided to you). Week Two : ( room 3E21 ) Samples will be thawed, and the quality of the product assessed. 2

1. The day prior to your lab period, the laboratory instructor will move your bagged samples from the freezer to the +1˚C cooler for thawing. 2. For the thawed samples, record your observations on colour and texture, and measure drip loss by remove the thawed product and re-weigh the bag with the damp paper towel in it (W damp ); the weight gained from the package is the drip loss. (Use that paper towel to blot any free water on the samples if needed prior to removal.) You can also taste the samples if you want. 3. You NEED to obtain data from all the groups for the two samples and two treatments. 4. Obtain freezing data for the single strength iced tea, and concentrated iced tea for graphing and discussion. (Will be posted on Canvas) Additional freezing of potato products may be done. Record and calculate your data here: Week # 1 Bag # 1 2 3 4 Sample name Straw Straw Mush Mush Treatment (Slow/medium) Med Slow Med Slow Weight of Bag/paper towel (W pack ) 8.57 9.22 8.40 8.10 Weight with sample (W pack+sample ) 108.49 108.94 107.49 108.24 Weight of sample (W sample ) (W pack+sample - W pack ) 99.20 99.72 99.09 100.14 Week # 2 Weight of damp pack (without sample) (W damp ) 28.37 26.28 23.97 27.46 Weight of drip loss (W drip ) (W damp - W pack ) 19.80 17.06 15.57 19.36 % Weight loss (W drip *100/ W sample ) 19.9% 17.1% 15.7% 19.3% Colour Red Red, some darkening Dark brown Dark brown, darker spots Texture Firmer, still softer than fresh Quite soft, still some firmness Semi firm Quite soft and mushy, slimy 3

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Questions: 1. In table format, compile initial sample weight, weight (drip) loss, %weight loss, and record observations on the colour and texture of the strawberry, and mushroom samples. Calculate %weight loss. You should have observed a greater exudate yield from slowly frozen strawberries, and mushrooms, and the more intense browning of slowly frozen mushrooms as compared to the other treatments. Explain this difference. (10 marks) Our observed data shows a greater exudate yield from the slowly frozen strawberries, but not the mushrooms. Error in our methodology may have caused this unexpected result. When comparing the data from the other groups, it is true that the mushrooms should also exude a greater yield when slowly frozen. The slowly frozen strawberries and mushrooms exude a greater drip loss yield because of ice crystal formation. When slowly frozen, the more gradual decrease in temperature, causing larger ice crystals. In the blast freezer, the temperature is rapidly dropped, forming smaller ice crystals. The large ice crystals break structures and cell walls, potentially causing rupturing. The rupturing releases water and other cell contents when thawed, causing more drip loss. It was observed that the slowly frozen mushrooms and strawberries appeared browner and slightly darker in colour than the fast frozen. This is enzymatic browning and is caused by polyphenol oxidase and peroxidase (Bro, Hanne, 1996). These enzymes catalyze the oxidation of compounds in the food, resulting in brown pigments. The slow freezing allows more time for these enzymes to interact with oxygen and turn brown. The fast frozen strawberries and mushrooms have a minimal amount of time that the enzymes can be active. 2. Plot the time-temperature data obtained for the iced tea and concentrated iced tea samples at slow and medium freezing rates. Calculate initial cooling rates ( o C/min) (before sample frozen) of each treatment, what conclusion you can make, and why? (5 marks) Initial Cooling Rates Regular Slow - -0.24 Concentrated Slow – -0.25 Regular Medium – -0.57 Concentrated Medium - -0.42 3. What and when were the initial freezing points of the iced tea products measured at slow and medium freezing rates? (The initial freezing point is the point on the graph where product temperature is not decreasing temporarily due to the phase change.) Explain any differences in initial freezing points at the slow and medium freezing rates and between products. Provide a more comprehensive definition for initial freezing point. What phenomenon appears to be occurring when the temperature rises in the time-temperature plots? What were the supercool temperatures reached for each treatment? (10 marks) Initial freezing points Regular Slow - -1.22 Concentrated Slow – -7.59 Regular Medium – -1.77 Concentrated Medium - -11.74 4

The initial freezing point for the slow freezing rate is lower than the initial freezing point for the medium freezing rate. The point is lower because the rate is more gradual, as it takes more time to freeze the substance. The concentrated iced tea has a lower freezing point for both freezing methods. When there is a solute dissolved in the solvent, the freezing point will be different than that of a pure solvent. In this case, iced tea dissolved in water. The concentrated iced tea has more solutes dissolved in it, meaning less water, thus lowering the freezing point more than the regular. The initial freezing point is when crystallization begins to occur, not necessarily when the product first drops below 0. This is when the phase change occurs. Supercooling happens when a liquid is cooled below its freezing point without crystalizing or freezing (Stonehouse, Evans, 2015) When the iced tea reaches the freezing point, it undergoes a phase change from a liquid to solid, and the temperature remains fairly constant. It is the release of latent heat during this phase change that keeps it in a thermodynamic equilibrium (constant) state (Stonehouse, Evans 2015). 4. Plot the time-temperature data for the potato freezing experiment using the provided data. The temperature of the freezing medium (brine solution) should also be shown as a line on the graph Why was the freezing point of the potato not at 0.0°C, and what and when was the freezing point of the potato? Calculate the cooling rate of the potato (°C per min) from 15°C to 0°C; -2°C to - 2.5°C; and -4 to -8°C. Discuss the factors influencing the cooling rate of the potato sample at each time interval, i.e., we’d expect the cooling rate of ice to be faster than liquid water but that may not have happened (10 marks) Freezing point – Cooling rate 15°C to 0°C = -15/274min = -0.055°C per min 0°C; -2°C = -2/756 = -0.9926°C per min -2°C to -2.5°C = -0.5/356 = -0.0014°C per min -4 to -8°C = -4/256 = 0.016°C per min Bonus The sodium chloride brine used showed a large freezing point depression. What would be the freezing point depression achieved by adding 35 grams of sodium chloride to 200 grams of water? Show your work. (2 marks). Upload typed answers to LAB QUESTIONS through Canvas no later than October 18 (one week after your lab) https://www-sciencedirect-com.cyber.usask.ca/science/article/pii/0169743996000196?via%3Dihub 5