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Chapter 10: Problem 24

Cake mixes and other packaged foods that require cooking often contain special directions for use at high elevations. Typically these directions indicate that the food should be cooked longer above \(5000 \mathrm{ft}\). Explain why it takes longer to cook something at higher elevations.

Short Answer

Expert verified
At higher elevations, atmospheric pressure decreases, causing the boiling point of water to decrease as well. This lower boiling point affects heat transfer through water during cooking processes like boiling, steaming, or simmering, resulting in foods taking longer to reach their desired internal temperature or become tenderized. Therefore, packaged food directions suggest longer cooking times at high elevations (above 5000 feet) to ensure that the food is cooked thoroughly and safely.

Step by step solution

01

Understand the relationship between atmospheric pressure and elevation

As elevation increases, atmospheric pressure decreases due to the decreasing weight of the air column above the given elevation. This reduced pressure has a direct effect on the boiling point of water.
02

Learn the effect of atmospheric pressure on boiling point of water

At sea-level, water boils at 100°C (212°F) because the atmospheric pressure is sufficient to keep water molecules contained in their liquid state until they gain enough energy to vaporize. However, at higher elevations where atmospheric pressure is lower, water molecules can escape into the vapor phase more quickly, requiring less energy. As a result, the boiling point of water decreases with increasing elevation.
03

Discover the implications of a lower boiling point on cooking time

Since cooking often involves the transfer of heat through water, a lower boiling point of water at high elevations means that water will boil at a lower temperature compared to sea-level. Cooking processes like boiling, steaming, or simmering rely on this heat transfer to break down the connective tissues in meat, cook vegetables to a tender state, and kill bacteria. Due to the decreased temperature at which water boils at high elevations, foods will take longer to reach their desired internal temperature or become tenderized.
04

Apply this understanding to packaged food directions

Packaged food directions for use at high elevations (above 5000 feet) often require longer cooking times to ensure that the food is cooked thoroughly since it can take longer at the lower boiling point to achieve the necessary internal temperature. These instructions are designed to adjust for the difference in boiling point and ensure the food is cooked correctly and safely. To conclude, it takes longer to cook something at higher elevations, primarily due to the decrease in atmospheric pressure and the resulting lower boiling point of water. This lower boiling point affects heat transfer and causes foods to require more time to cook thoroughly.

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Most popular questions from this chapter

Predict which substance in each of the following pairs would have the greater intermolecular forces. a. \(\mathrm{CO}_{2}\) or \(\mathrm{OCS}\) b. \(\mathrm{SeO}_{2}\) or \(\mathrm{SO}_{2}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) d. \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) or \(\mathrm{H}_{2} \mathrm{CO}\) e. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{H}_{2} \mathrm{CO}\)

The molar heat of fusion of sodium metal is \(2.60 \mathrm{~kJ} / \mathrm{mol}\), whereas its heat of vaporization is \(97.0 \mathrm{~kJ} / \mathrm{mol}\). a. Why is the heat of vaporization so much larger than the heat of fusion? b. What quantity of heat would be needed to melt \(1.00 \mathrm{~g}\) sodium at its normal melting point? c. What quantity of heat would be needed to vaporize \(1.00 \mathrm{~g}\) sodium at its normal boiling point? d. What quantity of heat would be evolved if \(1.00 \mathrm{~g}\) sodium vapor condensed at its normal boiling point?

The \(\mathrm{CsCl}\) structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 67 ). Given that the density of cesium chloride is \(3.97 \mathrm{~g} / \mathrm{cm}^{3}\), and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm}\), and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

The compounds \(\mathrm{Na}_{2} \mathrm{O}, \mathrm{CdS}\), and \(\mathrm{Zr} \mathrm{I}_{4}\) all can be described as cubic closest packed anions with the cations in tetrahedral holes. What fraction of the tetrahedral holes is occupied for each case?

Consider the following melting point data: $$ \begin{array}{|llllllll|} \hline \text { Compound: } & \mathrm{NaCl} & \mathrm{MgCl}_{2} & \mathrm{AlCl}_{3} & \mathrm{SiCl}_{4} & \mathrm{PCl}_{3} & \mathrm{SCl}_{2} & \mathrm{Cl}_{2} \\ \operatorname{mp}\left({ }^{\circ} \mathrm{C}\right): & 801 & 708 & 190 & -70 & -91 & -78 & -101 \\ \text { Compound: } & \mathrm{NaF} & \mathrm{MgF}_{2} & \mathrm{AlF}_{3} & \mathrm{SiF}_{4} & \mathrm{PF}_{5} & \mathrm{SF}_{6} & \mathrm{~F}_{2} \\ \operatorname{mp}\left({ }^{\circ} \mathrm{C}\right): & 997 & 1396 & 1040 & -90 & -94 & -56 & -220 \\ \hline \end{array} $$ Account for the trends in melting points in terms of interparticle forces.
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