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Chapter 6: Problem 8

Explain why oceanfront areas generally have smaller temperature fluctuations than inland areas.

Short Answer

Expert verified
Oceanfront areas experience smaller temperature fluctuations than inland areas due to the high specific heat capacity of water, which allows the ocean to absorb and store heat without significant changes in temperature. This results in milder temperatures for coastal areas with cooler summers and warmer winters. The ocean also redistributes heat, further stabilizing oceanfront temperatures. In contrast, inland areas have greater temperature fluctuations due to land materials' lower specific heat capacity and the absence of a large body of water to absorb and redistribute heat, leading to the "continental effect" of temperature extremes.

Step by step solution

01

Understand specific heat capacity

Specific heat capacity is a property of a substance that indicates the amount of heat required to change its temperature by one degree Celsius. Higher specific heat capacity means that the substance requires more heat to increase its temperature. Water, which covers oceanfront areas, has a high specific heat capacity compared to land materials, which cover inland areas.
02

Ocean's heat absorption capacity

Due to water's high specific heat capacity, the ocean can absorb and store a large amount of heat energy without significantly increasing its temperature. This means that the ocean takes longer to heat up during the day and longer to cool down at night. As a result, the temperature of the ocean remains relatively stable throughout the day and night.
03

Land's heat absorption capacity

In contrast, land materials have a lower specific heat capacity compared to water. This means that the land heats up and cools down more quickly than the ocean. Thus, land experiences more significant temperature fluctuations throughout the day and night.
04

Influence of the ocean on coastal climate

Oceanfront areas are heavily influenced by the ocean's stable temperature due to their proximity to the vast body of water. The ocean's ability to store heat results in milder temperatures for coastal areas, with cooler summers and warmer winters than areas further inland. The constant movement of the ocean also redistributes heat, ensuring that oceanfront temperatures remain more stable.
05

Continental effect on inland climate

Inland areas, on the other hand, are far away from the moderating influence of the ocean. These areas generally experience more significant temperature fluctuations, with hotter summers and colder winters. Since there is no large body of water nearby to help absorb and redistribute heat, inland areas endure temperature extremes known as the "continental effect." In conclusion, oceanfront areas generally have smaller temperature fluctuations than inland areas because of the ocean's high specific heat capacity and its ability to store and redistribute heat. This moderating effect creates a more stable climate for coastal regions compared to inland areas that experience greater temperature fluctuations due to the land's lower specific heat capacity and the continental effect.

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Most popular questions from this chapter

A swimming pool, \(10.0 \mathrm{~m}\) by \(4.0 \mathrm{~m}\), is filled with water to a depth of \(3.0 \mathrm{~m}\) at a temperature of \(20.2^{\circ} \mathrm{C}\). How much energy is required to raise the temperature of the water to \(24.6^{\circ} \mathrm{C} ?\)

Write reactions for which the enthalpy change will be a. \(\Delta H_{\mathrm{f}}^{\circ}\) for solid aluminum oxide. b. the standard enthalpy of combustion of liquid ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) .\) c. the standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid. d. \(\Delta H_{\mathrm{f}}^{\circ}\) for gaseous vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(\mathrm{g})\). e. the enthalpy of combustion of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\). f. the enthalpy of solution of solid ammonium bromide.

You have a 1.00-mole sample of water at \(-30 .{ }^{\circ} \mathrm{C}\) and you heat it until you have gaseous water at \(140 .{ }^{\circ} \mathrm{C}\). Calculate \(q\) for the entire process. Use the following data. Specific heat capacity of ice \(=2.03 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of water \(=4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of steam \(=2.02 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\text {fusion }}=6.02 \mathrm{~kJ} / \mathrm{mol}\left(\right.\) at \(\left.0^{\circ} \mathrm{C}\right)\) \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\text {vaporization }}=40.7 \mathrm{~kJ} / \mathrm{mol}\left(\right.\) at \(\left.100 .^{\circ} \mathrm{C}\right)\)

On Easter Sunday, April 3,1983, nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: \(2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \% \mathrm{HNO}_{3}\) by mass with a density of \(1.42 \mathrm{~g} / \mathrm{cm}^{3} .\) What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\left.\mathrm{NaNO}_{3}(a q)=-467 \mathrm{~kJ} / \mathrm{mol}\right)\) b. According to The Denver Post for April 4,1983 , authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\), what was their major concern?

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5 \mathrm{~L}\). Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.
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