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Chapter 12: Problem 115

It is not uncommon for the night sky temperature in desert regions to drop to \(-40^{\circ} \mathrm{C}\). If the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefticient for still air conditions is approximately \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), can a shallow pan of water freeze?

Short Answer

Expert verified
The difference in temperature between the water and the night sky temperature is \(60^{\circ}\mathrm{C}\). The rate of heat transfer, \(q\), is positive, meaning the water is losing heat to the surrounding air. The temperature difference is sufficient for the water to potentially freeze as long as the heat transfer is faster than the rise in temperature due to external heat sources. However, factors such as evaporation and imperfect isolation from other heat sources may affect this outcome. As a rough estimate, it is possible for the water in a shallow pan to freeze under the given conditions.

Step by step solution

01

Determine the difference in temperature

First, we need to find the difference in temperature between the water and the ambient air. Assuming the water is initially at the ambient air temperature, \(T_{water} = 20^{\circ}\mathrm{C}\) and the night sky temperature is \(T_{sky} = -40^{\circ}\mathrm{C}\), the difference in temperature, 螖T, can be calculated as: 螖T = \(T_{water} - T_{sky}\) 螖T = \(20 - (-40)\) 螖T = \(60^{\circ}\mathrm{C}\)
02

Calculate the rate of heat transfer

Next, we need to calculate the rate of heat transfer, \(q\), between the water and the surrounding air by following Newton's Law of cooling: \(q = h \cdot A \cdot (T_{water} - T_{ambient})\) where \(h\) is the convection coefficient for still air conditions, given as \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(A\) is the surface area of the water pan which is exposed to the air, \(T_{water}\) is the initial temperature of the water, and \(T_{ambient}\) is the ambient air temperature. The surface area, \(A\), is not given in the problem, but since we wish to determine if the water can freeze, let's analyze if the rate of heat transfer is positive or negative to help us know if the water is losing or gaining heat. Since the convection coefficient, \(h\), and surface area, \(A\), are both positive, we have: \(q = h \cdot A \cdot (T_{water} - T_{ambient}) > 0\) This means the water is losing heat to the surrounding air.
03

Determine if the water can freeze

Now, we need to assess if the difference in temperature is sufficient for the water to freeze. The freezing point of water is \(0^{\circ}\mathrm{C}\). Since the water is losing heat to the surrounding air, we need to check if the temperature difference of \(60^{\circ}\mathrm{C}\) is enough to cause the water's temperature to drop below the freezing point. In this case, the difference is sufficient for the water to freeze as long as the heat transfer is faster than the rise in temperature due to external heat sources (e.g., sunlight). However, in reality, certain factors such as evaporation and imperfect isolation from other heat sources make it difficult to definitively conclude that the water will freeze based on this analysis alone. We could further explore the factors affecting heat transfer and better estimate the time it would take for the water to freeze, but as a rough estimate, it is possible for the water in a shallow pan to freeze under the given conditions.

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Key Concepts

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

Newton's Law of Cooling
Understanding Newton's Law of Cooling is essential when delving into the principles of heat transfer in thermodynamics. Simply put, this law describes how the temperature of an object changes when it's exposed to an ambient temperature. The formula encapsulated by this law is given by:
\( q = h \cdot A \cdot (T_{object} - T_{ambient}) \).
In this equation, \( q \) represents the heat transfer per unit time, \( h \) stands for the convection coefficient, \( A \) is the surface area through which the heat transfer takes place, \( T_{object} \) indicates the temperature of the object, and \( T_{ambient} \) is the ambient temperature. The law implies that the rate of heat loss of the object is proportional to the difference in temperature between the object and its surroundings.
  • The greater this temperature difference, the faster the rate of heat transfer.
  • If the temperature of the object is higher than the ambient, the object cools down, and conversely, if it's lower, the object heats up.
  • Parameters like the convection coefficient play a critical role in determining how quickly the object can reach equilibrium with its environment.
This law can be applied to a multitude of situations, from understanding how a cup of tea cools down to analyzing the thermal dynamics of entire ecosystems.
Convection Coefficient
The convection coefficient, often denoted as \( h \), is a crucial factor in the heat transfer process. It quantifies the heat transfer between a solid surface and a fluid (such as air or water) moving past it. Its units are typically \( \text{W/m}^2 \cdot \text{K} \) (watts per square meter per degree Kelvin鈥攐r Celsius, as the degree size is the same for both scales).

Factors Influencing the Convection Coefficient


Several factors can affect the convection coefficient, including:
  • The properties of the fluid, such as viscosity, thermal conductivity, and density.
  • The velocity of the fluid flow.
  • The nature of the flow, which can be laminar (smooth) or turbulent (chaotic).
  • The geometry and surface roughness of the solid.
A higher convection coefficient indicates more efficient heat transfer. In the context of our exercise, the convection coefficient for still air (\(5 \text{W/m}^2 \cdot \text{K} \)) is relatively low compared to forced air conditions, which means heat transfer is less efficient in still air, but still considerable over time.
Freezing Point of Water
The freezing point of water is a constant physical property, set at \(0^\circ\text{C}\) (32掳F), providing a reference point for various thermal calculations and real-life applications. When water reaches this temperature, it undergoes a phase change from liquid to solid, assuming standard atmospheric pressure. Precisely knowing the freezing point is crucial when we evaluate the potential for water to freeze under given conditions.

Conditions Affecting the Freezing Point


For instance, in desert regions where the night temperature could sharply fall, as in our exercise scenario, even though the ambient temperature starts at \(20^\circ\text{C}\), the significant temperature differential can lead to heat being lost from the water, increasing the chances of reaching the freezing point.
  • Presence of impurities or solutes can lower the freezing point, a concept known as freezing point depression.
  • Pressure changes can also affect the freezing point slightly, though this is less significant for most surface-level applications.
  • The amount of water and the characteristics of its container can influence how quickly the water cools down and reaches the freezing temperature.
Understanding the freezing point in the context of heat transfer gives us insights into whether certain conditions, like the ones outlined in the exercise, would allow for water to transition into ice.

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Most popular questions from this chapter

A furnace with an aperture of 20 -mm diameter and emissive power of \(3.72 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\) is used to calibrate a heat flux gage having a sensitive area of \(1.6 \times 10^{-5} \mathrm{~m}^{2}\). (a) At what distance, measured along a normal from the aperture, should the gage be positioned to receive irradiation of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) ? (b) If the gage is tilted off normal by \(20^{\circ}\), what will be its irradiation? (c) For tilt angles of 0,20 , and \(60^{\circ}\), plot the gage irradiation as a function of the separation distance for values ranging from 100 to \(300 \mathrm{~mm}\).

A proposed method for generating electricity from solar irradiation is to concentrate the irradiation into a cavity that is placed within a large container of a salt with a high melting temperature. If all heat losses are neglected, part of the solar irradiation entering the cavity is used to melt the salt while the remainder is used to power a Rankine cycle. (The salt is melted during the day and is resolidified at night in order to generate electricity around the clock.) Consider conditions for which the solar power entering the cavity is \(q_{\mathrm{sal}}=7.50 \mathrm{MW}\) and the time rate of change of energy stored in the salt is \(\dot{E}_{\mathrm{st}}=3.45 \mathrm{MW}\). For a cavity opening of diameter \(D_{s}=1 \mathrm{~m}\), determine the heat transfer to the Rankine cycle, \(q_{R}\). The temperature of the salt is maintained at its melting point, \(T_{\text {salt }}=T_{\text {m }}=1000^{\circ} \mathrm{C}\). Neglect heat loss by convection and irradiation from the surroundings.

Isothermal furnaces with small apertures approximating a blackbody are frequently used to calibrate heat flux gages, radiation thermometers, and other radiometric devices. In such applications, it is necessary to control power to the furnace such that the variation of temperature and the spectral intensity of the aperture are within desired limits. (a) By considering the Planck spectral distribution, Equation \(12.30\), show that the ratio of the fractional change in the spectral intensity to the fractional change in the temperature of the furnace has the form $$ \frac{d I_{\lambda} / I_{\lambda}}{d T / T}=\frac{C_{2}}{\lambda T} \frac{1}{1-\exp \left(-C_{2} / \lambda T\right)} $$ (b) Using this relation, determine the allowable variation in temperature of the furnace operating at \(2000 \mathrm{~K}\) to ensure that the spectral intensity at \(0.65 \mu \mathrm{m}\) will not vary by more than \(0.5 \%\). What is the allowable variation at \(10 \mu \mathrm{m}\) ?

Consider the metallic surface of Example 12.7. Additional measurements of the spectral, hemispherical emissivity yield a spectral distribution which may be approximated as follows: (a) Determine corresponding values of the total, hemispherical emissivity \(\varepsilon\) and the total emissive power \(E\) at \(2000 \mathrm{~K}\). (b) Plot the emissivity as a function of temperature for \(500 \leq T \leq 3000 \mathrm{~K}\). Explain the variation.

A thermocouple whose surface is diffuse and gray with an emissivity of \(0.6\) indicates a temperature of \(180^{\circ} \mathrm{C}\) when used to measure the temperature of a gas flowing through a large duct whose walls have an emissivity of \(0.85\) and a uniform temperature of \(450^{\circ} \mathrm{C}\). (a) If the convection heat transfer coefficient between the thermocouple and the gas stream is \(h=125 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and there are negligible conduction losses from the thermocouple, determine the temperature of the gas. (b) Consider a gas temperature of \(125^{\circ} \mathrm{C}\). Compute and plot the thermocouple measurement error as a function of the convection coefficient for \(10 \leq 5\) \(h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What are the implications of your results?
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