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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Hydrogen can cause fire hazards, and hydrogen gas leaking into surrounding air can lead to spontaneous ignition with extremely hot flames. Even at very low leakage rate, hydrogen can sustain combustion causing extended fire damages. Hydrogen gas is lighter than air, so if a leakage occurs it accumulates under roofs and forms explosive hazards. To prevent such hazards, buildings containing source of hydrogen must have adequate ventilation system and hydrogen sensors. Consider a metal spherical vessel, with an inner diameter of \(5 \mathrm{~m}\) and a thickness of \(3 \mathrm{~mm}\), containing hydrogen gas at \(2000 \mathrm{kPa}\). The vessel is situated in a room with atmospheric air at \(1 \mathrm{~atm}\). The ventilation system for the room is capable of keeping the air fresh, provided that the rate of hydrogen leakage is below \(5 \mu \mathrm{g} / \mathrm{s}\). If the diffusion coefficient and solubility of hydrogen \(\mathrm{gas}\) in the metal vessel are \(1.5 \times 10^{-12} \mathrm{~m}^{2} / \mathrm{s}\) and \(0.005 \mathrm{kmol} / \mathrm{m}^{3}\).bar, respectively, determine whether or not the vessel is safely containing the hydrogen gas.

A thick wall made of natural rubber is exposed to pure oxygen gas on one side of its surface. Both the wall and oxygen gas are isothermal at \(25^{\circ} \mathrm{C}\), and the oxygen concentration at the wall surface is constant. Determine the time required for the oxygen concentration at \(x=5 \mathrm{~mm}\) to reach \(5 \%\) of its concentration at the wall surface.

Define the penetration depth for mass transfer, and explain how it can be determined at a specified time when the diffusion coefficient is known.

For the absorption of a gas (like carbon dioxide) into a liquid (like water) Henry's law states that partial pressure of the gas is proportional to the mole fraction of the gas in the liquid-gas solution with the constant of proportionality being Henry's constant. A bottle of soda pop \(\left(\mathrm{CO}_{2}-\mathrm{H}_{2} \mathrm{O}\right)\) at room temperature has a Henry's constant of \(17,100 \mathrm{kPa}\). If the pressure in this bottle is \(120 \mathrm{kPa}\) and the partial pressure of the water vapor in the gas volume at the top of the bottle is neglected, the concentration of the \(\mathrm{CO}_{2}\) in the liquid \(\mathrm{H}_{2} \mathrm{O}\) is (a) \(0.003 \mathrm{~mol}-\mathrm{CO}_{2} / \mathrm{mol}\) (b) \(0.007 \mathrm{~mol}-\mathrm{CO}_{2} / \mathrm{mol}\) (c) \(0.013 \mathrm{~mol}-\mathrm{CO}_{2} / \mathrm{mol}\) (d) \(0.022 \mathrm{~mol}-\mathrm{CO}_{2} / \mathrm{mol}\) (e) \(0.047 \mathrm{~mol}-\mathrm{CO}_{2} / \mathrm{mol}\)

The average heat transfer coefficient for air flow over an odd-shaped body is to be determined by mass transfer measurements and using the Chilton-Colburn analogy between heat and mass transfer. The experiment is conducted by blowing dry air at \(1 \mathrm{~atm}\) at a free stream velocity of \(2 \mathrm{~m} / \mathrm{s}\) over a body covered with a layer of naphthalene. The surface area of the body is \(0.75 \mathrm{~m}^{2}\), and it is observed that \(100 \mathrm{~g}\) of naphthalene has sublimated in \(45 \mathrm{~min}\). During the experiment, both the body and the air were kept at \(25^{\circ} \mathrm{C}\), at which the vapor pressure and mass diffusivity of naphthalene are \(11 \mathrm{~Pa}\) and \(D_{A B}=0.61 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. Determine the heat transfer coefficient under the same flow conditions over the same geometry.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free