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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
Short Answer: Yes, the claim can be considered reasonable if the reaction rate at -23°C is roughly half as fast as the reaction rate at -12°C, as determined by the Arrhenius equation. However, this analysis assumes that all other factors (e.g., storage conditions, humidity) remain constant and only the temperature affects the reaction rate. In real-world scenarios, there may be other factors that impact the storage duration of beef.

Step by step solution

01

Understand the relationship between temperature and reaction rate

According to the Arrhenius equation, the rate of a chemical reaction depends on the temperature, where the higher the temperature, the faster the reaction rate. The equation is as follows: k = Ae^(-Ea/RT) where k is the reaction rate, A is the pre-exponential factor, Ea is the activation energy, R is the universal gas constant, and T is the temperature in Kelvin.
02

Convert temperatures to Kelvin

To use the Arrhenius equation, we need to convert the temperatures from Celsius to Kelvin. The formula for conversion is: K = °C + 273.15 For -23°C: T1 = -23 + 273.15 = 250.15 K For -12°C: T2 = -12 + 273.15 = 261.15 K
03

Calculate the ratio of reaction rates at the two temperatures

Now, we will compare the rate of the reactions at the two temperatures by calculating the ratio of the reaction rates, assuming that the activation energy and the pre-exponential factor remain constant. We can write the ratio as follows: k1/k2 = e^(Ea/RT2 - Ea/RT1)
04

Substitute temperature values and simplify

By substituting the temperature values in the above equation, we get: k1/k2 = e^(Ea/(R*261.15) - Ea/(R*250.15)) Since k1 is the reaction rate at -23°C and k2 is the reaction rate at -12°C, the ratio k1/k2 represents the factor by which the reaction rate is slower at -23°C compared to -12°C.
05

Compare the calculated ratio with the claim

According to the claim, the beef can be stored for 2 years at -23°C but only for 1 year at -12°C. This implies that the reaction rate at -23°C should be roughly twice as slow as the reaction rate at -12°C (k1/k2 = 1/2). Let's simplify the ratio we obtained in Step 4: k1/k2 = e^(Ea/(R*261.15) - Ea/(R*250.15)) After evaluating this ratio, if the value of k1/k2 obtained is close to 1/2, then the claim appears to be reasonable. In conclusion, the claim can be considered reasonable if the reaction rate at -23°C is roughly half as fast as the reaction rate at -12°C, as determined by the Arrhenius equation. However, this analysis assumes that all other factors (e.g., storage conditions, humidity) remain constant and only the temperature affects the reaction rate. In real-world scenarios, there may be other factors that impact the storage duration of beef.

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Most popular questions from this chapter

Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite $\left(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\( and \)\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$. Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of $40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine the center temperature of each geometry after 10 , 20 , and \(60 \mathrm{~min}\). Solve this problem using the analytical oneterm approximation method.

A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete $\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.94 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right.\(, \)\rho=1600 \mathrm{~kg} / \mathrm{m}^{3}\(, and \)c_{p}=0.84 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ ) cooled to \(14^{\circ} \mathrm{C}\) during a cold night is heated again during the day by being exposed to ambient air at an average temperature of \(28^{\circ} \mathrm{C}\) with an average heat transfer coefficient of $14 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}$. Using the analytical one-term approximation method, determine \((a)\) how long it will take for the column surface temperature to rise to \(27^{\circ} \mathrm{C}\), (b) the amount of heat transfer until the center temperature reaches to \(28^{\circ} \mathrm{C}\), and \((c)\) the amount of heat transfer until the surface temperature reaches \(27^{\circ} \mathrm{C}\).

Carbon steel balls $\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}, k=64 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, \)\left.c_{p}=434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( initially at \)200^{\circ} \mathrm{C}\( are quenched in an oil bath at \)20^{\circ} \mathrm{C}$ for a period of \(3 \mathrm{~min}\). If the balls have a diameter of \(5 \mathrm{~cm}\) and the convection heat transfer coefficient is $450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, the center temperature of the balls after quenching will be (Hint: Check the Biot number.) (a) \(30.3^{\circ} \mathrm{C}\) (b) \(46.1^{\circ} \mathrm{C}\) (c) \(55.4^{\circ} \mathrm{C}\) (d) \(68.9^{\circ} \mathrm{C}\) (e) \(79.4^{\circ} \mathrm{C}\)

A large iron slab $\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\(, and \)k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ was initially heated to a uniform temperature of \(150^{\circ} \mathrm{C}\) and then placed on a concrete floor $\left(\rho=1600 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=840 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\(, and \)\left.k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)$. The concrete floor was initially at a uniform temperature of \(30^{\circ} \mathrm{C}\). Determine \((a)\) the surface temperature between the iron slab and concrete floor and \((b)\) the temperature of the concrete floor at the depth of \(25 \mathrm{~mm}\), if the surface temperature remains constant after \(15 \mathrm{~min}\).

Can the one-term approximate solutions for a plane wall exposed to convection on both sides be used for a plane wall with one side exposed to convection while the other side is insulated? Explain.

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