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Chapter 4: Problem 37

A body of water at \(0^{\circ} \mathrm{C}\) is subjected to a constant surface temperature of \(-10^{\circ} \mathrm{C}\) for 10 days. How thick is the surface layer of ice? Use \(L=320 \mathrm{~kJ} \mathrm{~kg}^{-1}\), \(k=2 \mathrm{~J} \mathrm{~m}^{-1} \mathrm{~s}^{-1} \mathrm{~K}^{-1}, c=4 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}, \rho=\) \(1000 \mathrm{~kg} \mathrm{~m}^{-3}\).

Short Answer

Expert verified
The ice thickness is approximately 10.39 meters.

Step by step solution

01

Understand the Problem

We have water at the freezing point with a constant temperature of -10°C applied on the surface. We need to find the thickness of ice formed after 10 days using the given constants: latent heat of fusion \(L = 320 \, \text{kJ} \cdot \text{kg}^{-1}\), thermal conductivity \(k = 2 \, \text{J} \cdot \text{m}^{-1} \cdot \text{s}^{-1} \cdot \text{K}^{-1}\), specific heat \(c = 4 \, \text{kJ} \cdot \text{kg}^{-1} \cdot \text{K}^{-1}\), and density \(\rho = 1000 \, \text{kg} \cdot \text{m}^{-3}\).
02

Expression for Ice Thickness

The equation for forming a layer of ice of thickness \(x\) under a constant negative temperature gradient over time \(t\) is given by:\[ x = \sqrt{\frac{2k(T_i - T_s)t}{\rho L}} \]Where:- \(T_i = 0^{\circ}C\) (initial temperature of water)- \(T_s = -10^{\circ}C\) (constant surface temperature)- \(t\) is the time in seconds.
03

Convert Time to Seconds

10 days must be converted to seconds for the calculation:\[ t = 10 \, \text{days} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour} = 864,000 \, \text{seconds} \]
04

Calculate Ice Thickness

Substitute the values into the expression derived in Step 2:\[ x = \sqrt{\frac{2 \times 2 \times (0 - (-10)) \times 864000}{1000 \times 320 \times 1000}} \]This simplifies to:\[ x = \sqrt{\frac{2 \times 2 \times 10 \times 864000}{320000}} \]Calculate:\[ x = \sqrt{\frac{34560000}{320000}} \]\[ x \approx \sqrt{108} \]\[ x \approx 10.39 \, \text{meters} \]
05

Conclusion

After 10 days of exposure to a constant surface temperature of \(-10^{\circ} \text{C}\), the ice formed will have a thickness of approximately 10.39 meters.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a crucial concept in understanding heat transfer, especially in processes like ice formation. It refers to the ability of a material to conduct heat. In this context, thermal conductivity ( k ) is used to determine how efficiently heat is transferred from the water to the surrounding air at different temperatures. A constant temperature is applied to the water surface, causing heat to flow from the water since it is warmer compared to the surrounding environment.

The formula for calculating the thickness of ice involves thermal conductivity because it determines the rate of heat loss through the water-ice boundary. The equation \( x = \sqrt{\frac{2k(T_i - T_s)t}{\rho L}} \) highlights thermal conductivity as foundational. The efficiency of this transfer impacts how fast ice forms, emphasizing thermal conductivity in practical scenarios like engineering and environmental science.
Latent Heat of Fusion
Latent heat of fusion plays a pivotal role in phase changes, such as freezing water to form ice. It is the amount of heat required to change a substance from a solid to a liquid state (or vice versa), without affecting the temperature. For this exercise, it is given as \( L = 320 \, \text{kJ} \, \text{kg}^{-1} \).

When water at the surface reaches \(0^{\circ} \text{C} \), it begins to convert into ice, but instead of changing temperature, energy is used to overcome the bonds between water molecules. The heat removed per kilogram of water to form ice is the latent heat of fusion. Without this concept, calculating the thickness of ice would be incomplete, since it accounts for the energy needed to change the water’s phase to solid ice.
Ice Formation
Ice formation occurs when water is cooled to \(0^{\circ} \text{C} \) or below and begins to solidify. The process begins with the surface layer freezing as it comes into contact with colder temperatures. In this situation, the constant surface temperature of \(-10^{\circ} \text{C} \) allows continuous heat removal from the water, leading to the gradual growth of the ice layer.

The thickness of the ice layer is influenced by several factors, including thermal conductivity, the temperature difference, and the latent heat of fusion. Each day, more water converts into ice as long as the environment remains cold enough to sustain the phase change. This controlled experiment calculation helps in predicting ice thickness accurately over a set period.
Phase Change
A phase change is a physical process where a substance transitions between different states of matter, like from liquid to solid, known as freezing. In this exercise, water at \(0^{\circ} \text{C} \) undergoes a phase change to ice due to the applied low temperature.

The equation for ice thickness considers the time ( t ) over which the temperature remains below freezing. The persistence of cold conditions results in significant ice formation within the timeframe. Phase change is vital, as it indicates the transformation of water's energy structure without altering temperature, directly impacting natural processes and engineered environments.
  • Importance in climate studies
  • Applications in refrigeration
  • Crucial in calculating the energy requirements for industrial processes

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Most popular questions from this chapter

Derive an expression for the temperature at the center of a planet of radius \(a\) with uniform density \(\rho\) and internal heat generation \(H .\) Heat transfer in the planet is by conduction only in the lithosphere, which extends from \(r=b\) to \(r=a\). For \(0 \leq r \leq b\) heat transfer is by convection, which maintains the temperature gradient \(d T / d r\) constant at the adiabatic value \(-\Gamma\). The surface temperature is \(T_{0}\). To solve for \(T(r),\) you need to assume that \(T\) and the heat flux are continuous at \(r=b\).

Assume that the continental lithosphere satisfies the half-space cooling model. If a continental region has an age of \(1.5 \times 10^{9}\) years, how much subsidence would have been expected to occur in the last 300 Ma? Take \(\rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}, \kappa=1 \mathrm{~mm}^{2} \mathrm{~s}^{-1}\), \(T_{m}-T_{0}=1300 \mathrm{~K},\) and \(\alpha_{v}=3 \times 10^{-5} \mathrm{~K}^{-1}\). Assume that the subsiding lithosphere is being covered to sea level with sediments of density \(\rho_{s}=2500 \mathrm{~kg} \mathrm{~m}^{-3}\).

Suppose that upon entering the Earth's atmosphere, the surface of a meteorite has been heated to the melting point and the molten material is carried away by the flow. It is of interest to calculate the rate at which melting removes material from the meteorite. For this purpose, consider the following problem. The surface of a semi-infinite half-space moves downward into the half-space with constant velocity \(V\), as indicated in Figure \(4.39 .\) The surface is always at the melting temperature \(T_{m},\) and melted material above the instantaneous surface is removed from the problem. Assume that the surface of the half-space is melted by a constant heat flux \(q_{m}\) into the half-space from above the surface. Assume also that far from the melting surface the temperature is \(T_{0}\); that is, \(T \rightarrow T_{0}\) as \(\zeta \rightarrow \infty\). Find the temperature distribution in the half-space as a function of time \(T(\zeta, t),\) and determine \(V\) in terms of \(q_{m}\) and the thermodynamic properties of the rock. Account for the latent heat \(L\) required to melt the material.

The exponential depth dependence of heat production is preferred because it is self-preserving upon erosion. However, many alternative models can be prescribed. Consider a two-layer model with \(H=\) \(H_{1}\) and \(k=k_{1}\) for \(0 \leq y \leq h_{1},\) and \(H=H_{2}\) and \(k=k_{2}\) for \(h_{1} \leq y \leq h_{2} .\) For \(y>h_{2}, H=0\) and the upward heat flux is \(q_{m} .\) Determine the surface heat flow and temperature at \(y=h_{2}\) for \(\rho_{1}=2600 \mathrm{~kg} \mathrm{~m}^{-3}\), \(\rho_{2}=3000 \mathrm{~kg} \mathrm{~m}^{-3}, k_{1}=k_{2}=2.4 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) \(h_{1}=8 \mathrm{~km}, h_{2}=40 \mathrm{~km}, \rho_{1} H_{1}=2 \mu W \mathrm{~m}^{-3}, \rho_{2} H_{2}=\) \(0.36 \mu W \mathrm{~m}^{-3}, T_{0}=0^{\circ} \mathrm{C},\) and \(q_{m}=28 \mathrm{~mW} \mathrm{~m}^{-2}\).

Estimate the effects of variations in bottom water temperature on measurements of oceanic heat flow by using the model of a semi-infinite half-space subjected to periodic surface temperature fluctuations. Such water temperature variations at a specific location on the ocean floor can be due to, for example, the transport of water with variable temperature past the site by deep ocean currents. Find the amplitude of water temperature variations that cause surface heat flux variations of \(40 \mathrm{~mW} \mathrm{~m}^{-2}\) above and below the mean on a time scale of 1 day. Assume that the thermal conductivity of sediments is \(0.8 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) and the sediment thermal diffusivity is \(0.2 \mathrm{~mm}^{2} \mathrm{~s}^{-1}\).
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