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It is claimed that beef can be stored for up to two years at \(-23^{\circ} \mathrm{C}\) but no more than one year at \(-12^{\circ} \mathrm{C}\). Is this claim reasonable? Explain.

Short Answer

Expert verified
Answer: Based on the Arrhenius equation and the relationship between temperature and reaction rates, it is reasonable to assume that the claim is plausible. Although we cannot determine the exact quantitative relationship without more information, the lower temperature of -23°C can potentially lead to a decreased reaction rate, allowing the beef to be stored longer than one year.

Step by step solution

01

Understanding the Arrhenius equation

The Arrhenius equation is used to demonstrate the relationship between the rate of a reaction and temperature. It is given as: k = Ae^(-Ea / RT) where: - k is the reaction rate constant - A is the pre-exponential factor (frequency factor) - Ea is the activation energy of the reaction - R is the gas constant (8.314 J/(mol·K)) - T is the temperature (in Kelvin) Note that since we're only comparing the relative rates, we don't need to explicitly calculate the pre-exponential factor.
02

Convert given temperatures to Kelvin

First, we need to convert the temperatures given in Celsius to Kelvin. The relationship between Celsius and Kelvin temperatures is: T(K) = T(°C) + 273.15 - For -23°C: T1 = -23 + 273.15 = 250.15 K - For -12°C: T2 = -12 + 273.15 = 261.15 K
03

Determine the ratio of reaction rates at both temperatures

Using the Arrhenius equation, we can establish a relationship between the reaction rates at both temperatures. Let k1 and k2 be the reaction rates at temperatures T1 and T2 respectively. The ratio of reaction rates (k2/k1) can be found by dividing the Arrhenius equations for both temperatures: (k2/k1) = e^(-(Ea / RT2) + (Ea / RT1)) Now, we don't know the values of Ea and R, but we can analyze the relationship between the temperatures and reaction rates.
04

Analyze the claim based on the ratio of reaction rates

Given that spoilage is a reaction, the claim is reasonable if the ratio of reaction rates (k2/k1) is equal to or greater than 2, which means that the beef spoils at least twice as fast at -12°C compared to -23°C. We cannot determine an exact value of the ratio without exact values for Ea and R, but we know that the exponential term inside the Arrhenius equation will decrease as the temperature decreases. Therefore, it is feasible that the ratio can be equal to or greater than 2, making the claim reasonable. In conclusion, although we cannot confirm the exact quantitative relationship between the reaction rates, it is reasonable to assume that the claim of beef lasting up to two years at -23°C compared to one year at -12°C is plausible based on the Arrhenius equation and the effect of temperature on reaction rates.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Rate Constant
When it comes to understanding how fast a chemical reaction can proceed, the reaction rate constant, denoted by the symbol k, is an essential quantity. The reaction rate constant is a measure of the speed at which reactants are converted into products under specific conditions, including temperature and pressure.

The value of k is not just a number to plug into equations; it encapsulates vital information about the reaction. For example, in a refrigeration scenario, the rate at which meat spoils is a reaction that we can describe using a rate constant. A higher reaction rate constant at a given temperature means that the spoilage reactions are happening faster. This constant is crucial for food industries where packaging, storage, and shelf life are tied closely to the speed of these reactions.

In the exercise, understanding k helps explain why beef may spoil slower at lower temperatures, indicating that k is lower when the beef is stored at -23°C compared to that at -12°C. It reflects the general rule that most chemical reactions, including spoilage, slow down as temperatures decrease.
Activation Energy
The term activation energy, symbolized by the variable Ea, refers to the minimum amount of energy required to initiate a chemical reaction. Imagine it as a barrier that reactants must overcome to transform into products. A higher activation energy means that fewer molecules have the necessary energy to react at a given temperature, which slows down the reaction rate.

In the context of food preservation, activation energy is the 'hill' that spoilage-causing bacteria or reactions need to climb to make the beef go bad. Lower temperatures make this 'hill' even harder to surmount, as fewer bacteria or molecules have the required energy, resulting in longer preservation times. This concept explains why there's a difference in shelf life for beef stored at the temperatures provided in the exercise—beef has a longer shelf life at -23°C because the activation energy barrier is more difficult to overcome at this lower temperature, leading to slower spoilage rates.
Temperature Conversion
The exercise provided us with temperatures in degrees Celsius, but in order to apply them in the Arrhenius equation, we must perform a temperature conversion to Kelvin. This is necessary because Kelvin is the standard unit of temperature in the scientific community, particularly in thermodynamics and kinetics calculations.

To convert from Celsius to Kelvin, we add 273.15 to the Celsius temperature. This conversion is based on the fact that 0 K, or absolute zero, is the point where molecules theoretically stop moving and is equivalent to -273.15°C. So, for the exercise, -23°C becomes 250.15 K and -12°C becomes 261.15 K. These Kelvin temperatures can then be used to calculate how the reaction rate constant changes with temperature, which is central to the question posed about the storage duration of beef at different temperatures. This calculation underscores the importance of precision in scientific measurements and translations between temperature scales.

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Most popular questions from this chapter

What are the environmental factors that affect the growth rate of microorganisms in foods?

Thick slabs of stainless steel \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(\left.3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and copper \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=117 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) are placed under an array of laser diodes, which supply an energy pulse of \(5 \times 10^{7} \mathrm{~J} / \mathrm{m}^{2}\) instantaneously at \(t=0\) to both materials. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Determine the temperatures of both slabs, at \(5 \mathrm{~cm}\) from the surface and \(60 \mathrm{~s}\) after receiving an energy pulse from the laser diodes.

The soil temperature in the upper layers of the earth varies with the variations in the atmospheric conditions. Before a cold front moves in, the earth at a location is initially at a uniform temperature of \(10^{\circ} \mathrm{C}\). Then the area is subjected to a temperature of \(-10^{\circ} \mathrm{C}\) and high winds that resulted in a convection heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) on the earth's surface for a period of \(10 \mathrm{~h}\). Taking the properties of the soil at that location to be \(k=0.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), determine the soil temperature at distances \(0,10,20\), and \(50 \mathrm{~cm}\) from the earth's surface at the end of this \(10-\mathrm{h}\) period.

In Betty Crocker's Cookbook, it is stated that it takes \(2 \mathrm{~h} \mathrm{} 45 \mathrm{~min}\) to roast a \(3.2-\mathrm{kg}\) rib initially at \(4.5^{\circ} \mathrm{C}\) "rare" in an oven maintained at \(163^{\circ} \mathrm{C}\). It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered rare done when the thermometer inserted into the center of the thickest part of the meat registers \(60^{\circ} \mathrm{C}\). The rib can be treated as a homogeneous spherical object with the properties \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(0.91 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Determine \((a)\) the heat transfer coefficient at the surface of the rib; \((b)\) the temperature of the outer surface of the rib when it is done; and \((c)\) the amount of heat transferred to the rib. \((d)\) Using the values obtained, predict how long it will take to roast this rib to "medium" level, which occurs when the innermost temperature of the rib reaches \(71^{\circ} \mathrm{C}\). Compare your result to the listed value of \(3 \mathrm{~h} \mathrm{} 20 \mathrm{~min}\). If the roast rib is to be set on the counter for about \(15 \mathrm{~min}\) before it is sliced, it is recommended that the rib be taken out of the oven when the thermometer registers about \(4^{\circ} \mathrm{C}\) below the indicated value because the rib will continue cooking even after it is taken out of the oven. Do you agree with this recommendation? Solve this problem using analytical one-term approximation method (not the Heisler charts).

Chickens with an average mass of \(1.7 \mathrm{~kg}(k=\) \(0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) initially at a uniform temperature of \(15^{\circ} \mathrm{C}\) are to be chilled in agitated brine at \(-7^{\circ} \mathrm{C}\). The average heat transfer coefficient between the chicken and the brine is determined experimentally to be \(440 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the average density of the chicken to be \(0.95 \mathrm{~g} / \mathrm{cm}^{3}\) and treating the chicken as a spherical lump, determine the center and the surface temperatures of the chicken in \(2 \mathrm{~h}\) and \(45 \mathrm{~min}\). Also, determine if any part of the chicken will freeze during this process. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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