Freeze-dried foods have been dehydrated in a vacuum. During the process, the food freezes and must be heated to facilitate dehydration. Explain both how the vacuum speeds up dehydration and why the food freezes as a result.

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Ice Water: A restaurant owner wanted to impress his guests with some very cold drinks. He bought a freezer that keeps ice at a constant temperature of -80 °C. In his first experiment, he added 541 g of water at a temperature of 56 °C into a large bucket of ice. When he checked back 30 minutes later, he found that the ice had not melted. Instead, all of the water had frozen! How much ice was in the bucket? Assume the experiment was carried out in a thermally insulated bucket and that the final temperature of the ice was 0° C. The specific heat of liquid water is 4.184 J/g.K and the specific heat of ice is 2.108 j/g.K. The heat of fusion of water is 333.55 J/g.

(5%) Problem 10: You read in a popular diet blog that the author claims you can lose up to half a pound of body fat per week if, instead of drinking 2.4 liters (i.e., 2.4 kilograms) of water every day, you can eat the same amount of water in the form of ice. The reasoning is that your body must first melt the ice and then bring the ice water up to the body temperature of 37°C, which is a process that requires a significant amount of energy at the expense of "burned" body fat. Thankfully, you just finished discussing heat and phase changes in your physics class, so you can calculate exactly how much energy is expended. The specific heat and latent heat of fusion for water are 4.186 × 10³ J/(kg·°C) and 3.34×105 J/kg, respectively. 50% Part (a) How many kilocalories do you expend when you eat 2.4 kg of ice at 0°C? Note that 1 kcal = 4186 J. Q=7174 * Attempts Remain 50% Part (b) Assuming the metabolism of one pound of body fat produces 3500 kcal of energy, how many pounds of body fat could…