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What happens when you spread salt crystals over an icy sidewalk? Why is this procedure rarely used in very cold climates?

Short Answer

Expert verified
Salt lowers the freezing point of ice, melting it, but is less effective below -9°C.

Step by step solution

01

Understanding the effect of salt on ice

When salt is spread over ice, it lowers the freezing point of water. This means that the ice can melt at a temperature lower than the usual 0°C (32°F). The salt dissolves in the thin layer of water on the surface, forming a saltwater solution that helps to melt the ice.
02

Chemical Explanation

The process of lowering the freezing point is called 'freezing point depression.' When salt (such as sodium chloride) dissolves in water, it dissociates into its components (Na^+ and Cl^−). These ions interfere with the formation of the solid ice structure, effectively lowering the temperature at which water can freeze.
03

Why it is less effective in very cold climates

In very cold climates, where temperatures are significantly below 0°C, the effectiveness of salt is reduced. Once the temperature drops below -9°C (15°F), the ability of typical road salt to lower the freezing point is limited, making it less effective in melting ice.
04

Alternative Methods

In very cold climates, alternatives like sand or specially formulated ice melts that work at lower temperatures are often used. These materials can provide traction and sometimes encourage melting even at lower temperatures than regular salt can handle.

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Key Concepts

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

Melting Ice with Salt
When temperatures drop below freezing, icy sidewalks and roads can become hazardous. One common way to combat this ice is by sprinkling salt, which serves as an effective agent for melting ice. This method leverages the concept of 'freezing point depression'. Here's how it works: normally, water freezes at 0°C (32°F).

However, when salt is introduced to ice, it dissolves into the thin layer of liquid water that is always present on the ice's surface. The resulting saltwater solution has a lower freezing point than pure water, enabling the ice to melt even in chilly conditions. In essence, the salt disrupts the orderly structure of ice, hindering its ability to remain solid until the temperatures drop even further.

Salt is readily available and cost-effective, making it a popular choice for road and sidewalk maintenance during winter storms.
Limitations of Road Salt
While salt is widely used to melt ice, it has its own set of limitations. The strategy of using salt becomes notably less effective as temperatures approach extreme lows. Most types of road salt, such as sodium chloride, are effective only down to approximately -9°C (15°F). Beyond this point, the salt can't sufficiently lower the freezing point of water, and hence, it does not work well in melting the ice.

Moreover, excessive use of salt can lead to environmental concerns. It can seep into soil and water sources, potentially harming plants, animals, and aquatic life. Additionally, prolonged exposure to salt can lead to corrosion of metals, which may cause damage to vehicles and infrastructure. These factors highlight the need for careful consideration when using salt, especially in areas prone to very cold climates.
Alternative Methods to Salt
In regions where temperatures routinely plummet well below freezing, there is a pressing need for more effective methods than traditional road salt. Some alternatives include:
  • Sand: While it does not melt the ice, sand provides traction, making surfaces less slippery.
  • Calcium chloride: This compound melts ice at much lower temperatures and is effective down to about -25°C (-13°F).
  • Magnesium chloride: Another alternative that works well at lower temperatures and is considered less corrosive than sodium chloride.
  • Brines: Liquid mixtures of salt with water that can be applied before a storm to prevent ice from bonding to surfaces.
These methods help maintain safety on walkways and roads while addressing the limitations and environmental impact of traditional salt usage. Employing a combination of techniques can often offer the best results in extreme winter conditions.

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Most popular questions from this chapter

An inventor proposes to make a heat engine using water/ice as the working substance, taking advantage of the fact that water expands as it freezes. A weight to be lifted is placed on top of a piston over a cylinder of water at \(1^{\circ} \mathrm{C}\). The system is then placed in thermal contact with a low-temperature reservoir at \(-1^{\circ} \mathrm{C}\) until the water freezes into ice, lifting the weight. The weight is then removed and the ice is melted by putting it in contact with a high-temperature reservoir at \(1^{\circ} \mathrm{C}\). The inventor is pleased with this device because it can seemingly perform an unlimited amount of work while absorbing only a finite amount of heat. Explain the flaw in the inventor's reasoning, and use the Clausius-Clapeyron relation to prove that the maximum efficiency of this engine is still given by the Carnot formula, \(1-T_{c} / T_{h}\)

How can diamond ever be more stable than graphite, when it has less entropy? Explain how at high pressures the conversion of graphite to diamond can increase the total entropy of the carbon plus its environment.

Graphite is more compressible than diamond. (a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text? (b) The isothermal compressibility of graphite is about \(3 \times 10^{-6} \mathrm{bar}^{-1},\) while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem \(1.46 .\) ) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).

The density of ice is \(917 \mathrm{kg} / \mathrm{m}^{3}.\) (a) Use the Clausius-Clapeyron relation to explain why the slope of the phase boundary between water and ice is negative. (b) How much pressure would you have to put on an ice cube to make it melt \(a t-1^{\circ} \mathrm{C} ?\) (c) Approximately how deep under a glacier would you have to be before the weight of the ice above gives the pressure you found in part (b)? (Note that the pressure can be greater at some locations, as where the glacier flows over a protruding rock.) (d) Make a rough estimate of the pressure under the blade of an ice skate, and calculate the melting temperature of ice at this pressure. Some authors have claimed that skaters glide with very little friction because the increased pressure under the blade melts the ice to create a thin layer of water. What do you think of this explanation?

A mixture of one part nitrogen and three parts hydrogen is heated, in the presence of a suitable catalyst, to a temperature of \(500^{\circ} \mathrm{C}\). What fraction of the nitrogen (atom for atom) is converted to ammonia, if the final total pressure is 400 atm? Pretend for simplicity that the gases behave ideally despite the very high pressure. The equilibrium constant at \(500^{\circ} \mathrm{C}\) is \(6.9 \times 10^{-5}\). (Hint: You'll have to solve a quadratic equation.)

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