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

The water main in the cities must be placed at sufficient depth below the earth's surface to avoid freezing during extended periods of subfreezing temperatures. Determine the minimum depth at which the water main must be placed at a location where the soil is initially at \(15^{\circ} \mathrm{C}\) and the earth's surface temperature under the worst conditions is expected to remain at \(-10^{\circ} \mathrm{C}\) for a period of 75 days. Take the properties of soil at that location to be \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.4 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Answer: \(7.05 \mathrm{~m}\)

A small chicken \(\left(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=0.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) can be approximated as an \(11.25\)-cm-diameter solid sphere. The chicken is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is to be cooked in an oven maintained at \(220^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). With this idealization, the temperature at the center of the chicken after a 90 -min period is (a) \(25^{\circ} \mathrm{C}\) (b) \(61^{\circ} \mathrm{C}\) (c) \(89^{\circ} \mathrm{C}\) (d) \(122^{\circ} \mathrm{C}\) (e) \(168^{\circ} \mathrm{C}\)

An experiment is to be conducted to determine heat transfer coefficient on the surfaces of tomatoes that are placed in cold water at \(7^{\circ} \mathrm{C}\). The tomatoes \((k=0.59 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=\) \(\left.0.141 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \rho=999 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.99 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) with an initial uniform temperature of \(30^{\circ} \mathrm{C}\) are spherical in shape with a diameter of \(8 \mathrm{~cm}\). After a period of 2 hours, the temperatures at the center and the surface of the tomatoes are measured to be \(10.0^{\circ} \mathrm{C}\) and \(7.1^{\circ} \mathrm{C}\), respectively. Using analytical one-term approximation method (not the Heisler charts), determine the heat transfer coefficient and the amount of heat transfer during this period if there are eight such tomatoes in water.

Consider a 7.6-cm-long and 3-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(\left.k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Fifteen such meat chunks initially at \(2^{\circ} \mathrm{C}\) are dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer during the first 8 minutes of cooking is (a) \(71 \mathrm{~kJ}\) (b) \(227 \mathrm{~kJ}\) (c) \(238 \mathrm{~kJ}\) \(\begin{array}{ll}\text { (d) } 269 \mathrm{~kJ} & \text { (e) } 307 \mathrm{~kJ}\end{array}\)

In a production facility, large plates made of stainless steel \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) of \(40 \mathrm{~cm}\) thickness are taken out of an oven at a uniform temperature of \(750^{\circ} \mathrm{C}\). The plates are placed in a water bath that is kept at a constant temperature of \(20^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The time it takes for the surface temperature of the plates to drop to \(100^{\circ} \mathrm{C}\) is (a) \(0.28 \mathrm{~h}\) (b) \(0.99 \mathrm{~h}\) (c) \(2.05 \mathrm{~h}\) (d) \(3.55 \mathrm{~h}\) (e) \(5.33 \mathrm{~h}\)

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