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Consider a shallow body of water. Is it possible for this water to freeze during a cold and dry night even when the ambient air and surrounding surface temperatures never drop to \(0^{\circ} \mathrm{C}\) ? Explain.

Short Answer

Expert verified
Explain your answer. Answer: Yes, it is possible for a shallow body of water to freeze under these conditions due to factors like the cooling effect of evaporation, the presence of impurities or dissolved salts, and the phenomenon of supercooling. These factors can cause water to freeze at temperatures slightly higher than the standard freezing point of \(0^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the problem

We are given a shallow body of water and the fact that the ambient air and surrounding surface temperatures never drop to \(0^{\circ} \mathrm{C}\). The goal is to determine if the water can still freeze under these conditions.
02

Consider factors that affect freezing point

The freezing point of water can be influenced by several factors, including: 1. Salt content: The presence of dissolved salts can lower the freezing point of water. This is known as freezing point depression. 2. Evaporation: Water can lose heat through evaporation, leading to a decrease in temperature. This is particularly relevant during a cold and dry night.
03

Evaluate the impact of evaporation on freezing

On a cold and dry night, the evaporation rate will likely be higher due to the lower humidity. Evaporation is a cooling process, as it removes heat from the water's surface, causing the temperature to decrease. In the case of a shallow body of water, the cooling effect of evaporation could affect the entire volume of water more quickly than in a larger body of water.
04

Explore exceptions to the \(0^{\circ} \mathrm{C}\) freezing point

While the standard freezing point of water is \(0^{\circ} \mathrm{C}\), it is affected by the aforementioned factors, like salt content and evaporation. Additionally, water can sometimes freeze at temperatures slightly higher than the standard freezing point due to the presence of impurities or the phenomenon of supercooling.
05

Conclude whether water can freeze under these conditions

Considering the factors that can affect the freezing point of water, it is possible for a shallow body of water to freeze during a cold and dry night even when the ambient air and surrounding surface temperatures never drop to \(0^{\circ} \mathrm{C}\). The cooling effect of evaporation, as well as any impurities in the water, can contribute to freezing even at slightly higher temperatures.

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Key Concepts

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

Phase Change
The transition of matter from one state to another, such as from liquid to solid, is known as a phase change. In the context of water, the phase change we are most interested in is freezing, where water turns into ice. This typically occurs at the freezing point of water, which is conventionally \(0^{\textdegree} \text{C}\) under standard atmospheric pressure. However, several factors can alter this temperature, a concept known as freezing point depression.

During a phase change, the temperature of the substance remains constant while the state change is in progress. This allows for the latent heat, or the hidden energy, to facilitate the change in state rather than altering the temperature. For water to freeze, it must release its latent heat to the environment, which can happen even if the ambient temperature is above \(0^{\textdegree} \text{C}\) under certain conditions, such as a cold and dry night with high rates of evaporation.
Evaporation Cooling
Evaporation cooling is a natural phenomenon where a liquid absorbs heat from its surroundings as it transitions into a gas, resulting in a cooler environment. This effect is commonly experienced when sweat evaporates from the skin, making us feel cooler on hot days.

Role in Freezing Water

On a cold, dry night, a shallow body of water can be significantly affected by evaporation cooling. As water molecules at the surface gain enough energy to escape into the air, they take heat with them, which reduces the temperature of the remaining liquid. This process can accelerate the cooling of the water body, potentially bringing its temperature below \(0^{\textdegree} \text{C}\) despite higher ambient temperatures, leading to freezing.

It is important to note that evaporation rate—and thus the cooling effect—is greater when the air is dry because dry air has a higher capacity to take in more water vapor. This is why the problem specified the conditions as 'cold and dry' night, which are ideal for rapid evaporation and subsequent cooling.
Supercooling
Supercooling is a state where a liquid remains liquid below its normal freezing point without becoming a solid. This often happens to pure water, which can be chilled well below \(0^{\textdegree} \text{C}\) without freezing, provided there are no impurities or disturbance to serve as nucleation sites for ice crystals to form.

Water Supercooling

Even in the absence of evaporation cooling, supercooling can occur in a shallow body of water. Factors contributing to supercooling include the absence of impurities and a perfectly still environment. It is a delicate balance because even slight disturbances can cause the supercooled water to suddenly freeze. Supercooling is a common reason why water might freeze at temperatures slightly above \(0^{\textdegree} \text{C}\) or not freeze at temperatures well below it until a disturbance triggers the rapid formation of ice.

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Most popular questions from this chapter

A glass bottle washing facility uses a well agi(Es) tated hot water bath at \(50^{\circ} \mathrm{C}\) with an open top that is placed on the ground. The bathtub is \(1 \mathrm{~m}\) high, \(2 \mathrm{~m}\) wide, and \(4 \mathrm{~m}\) long and is made of sheet metal so that the outer side surfaces are also at about \(50^{\circ} \mathrm{C}\). The bottles enter at a rate of 800 per minute at ambient temperature and leave at the water temperature. Each bottle has a mass of \(150 \mathrm{~g}\) and removes \(0.6 \mathrm{~g}\) of water as it leaves the bath wet. Makeup water is supplied at \(15^{\circ} \mathrm{C}\). If the average conditions in the plant are \(1 \mathrm{~atm}, 25^{\circ} \mathrm{C}\), and 50 percent relative humidity, and the average temperature of the surrounding surfaces is \(15^{\circ} \mathrm{C}\), determine (a) the amount of heat and water removed by the bottles themselves per second, \((b)\) the rate of heat loss from the top surface of the water bath by radiation, natural convection, and evaporation, \((c)\) the rate of heat loss from the side surfaces by natural convection and radiation, and \((d)\) the rate at which heat and water must be supplied to maintain steady operating conditions. Disregard heat loss through the bottom surface of the bath and take the emissivities of sheet metal and water to be \(0.61\) and \(0.95\), respectively.

Consider a thin layer of liquid water on a concrete surface. The surrounding air is dry with a convection heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The liquid water has an emissivity of \(0.95\), and the air and surrounding temperature is \(30^{\circ} \mathrm{C}\). If the layer of liquid water has a uniform temperature of \(20^{\circ} \mathrm{C}\), determine the conduction heat flux through the concrete.

Liquid methanol is accidentally spilt on a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) laboratory bench and covered the entire bench surface. A fan is providing a \(20 \mathrm{~m} / \mathrm{s}\) air flow parallel over the bench surface. The air is maintained at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), and the concentration of methanol in the free stream is negligible. If the methanol vapor at the air-methanol interface has a pressure of \(4000 \mathrm{~Pa}\) and a temperature of \(25^{\circ} \mathrm{C}\), determine the evaporation rate of methanol in molar basis.

A thin plastic membrane separates hydrogen from air. The molar concentrations of hydrogen in the membrane at the inner and outer surfaces are determined to be \(0.045\) and \(0.002 \mathrm{kmol} / \mathrm{m}^{3}\), respectively. The binary diffusion coefficient of hydrogen in plastic at the operation temperature is \(5.3 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). Determine the mass flow rate of hydrogen by diffusion through the membrane under steady conditions if the thickness of the membrane is (a) \(2 \mathrm{~mm}\) and (b) \(0.5 \mathrm{~mm}\).

In transient mass diffusion analysis, can we treat the diffusion of a solid into another solid of finite thickness (such as the diffusion of carbon into an ordinary steel component) as a diffusion process in a semi-infinite medium? Explain.

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