Question

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

The ice layer is approximately 23.2 cm thick.

Step-by-step solution

Understand the Problem

We are given a body of water at \(0^{\circ} \mathrm{C}\) and a surface temperature of \(-10^{\circ} \mathrm{C}\). The task is to find the thickness of the ice layer formed over 10 days. Key constants are provided: \(L = 320 \, \mathrm{kJ} \, \mathrm{kg}^{-1}\), \(k = 2 \, \mathrm{J} \, \mathrm{m}^{-1} \, \mathrm{s}^{-1} \, \mathrm{K}^{-1}\), \(c = 4 \, \mathrm{kJ} \, \mathrm{kg}^{-1} \, \mathrm{K}^{-1}\), and \(\rho = 1000 \mathrm{~kg} \, \mathrm{m}^{-3}\).

Establish Heat Transfer Equation

Use the heat conduction equation for freezing: \(Q = \frac{kA(T_s - T_i)t}{d}\). Here, \(Q\) is the heat lost, \(A\) is the area, \(T_s\) is the surface temperature, \(T_i\) is the initial temperature of water, \(t\) is the time, and \(d\) is the thickness of ice layer.

Relate Heat Lost to Ice Formation

The heat lost \(Q\) can also be expressed as \(Q = mL\), where \(m\) is the mass of ice, and \(L\) is the latent heat of fusion. The mass \(m\) is \(\rho \cdot A \cdot d\). Hence, \(Q = \rho A d L\).

Equate and Solve for Thickness

Equate the two expressions for \(Q\): \[ \frac{kA(T_s - T_i)t}{d} = \rho A d L \].Cancel \(A\) and solve for \(d\):\[ d^2 = \frac{k(T_s - T_i)t}{\rho L} \].Substitute \(T_s = -10\), \(T_i = 0\), \(t = 10 \times 24 \times 60 \times 60\) seconds:\[ d^2 = \frac{2 \times 10 \times 24 \times 3600}{1000 \times 320000} \].

Calculate the Thickness

Perform the calculation:\[ d^2 = \frac{2 \times 864000}{320000000} \approx 0.054 \].\[ d = \sqrt{0.054} \approx 0.232 \text{ meters or } 23.2 \text{ cm} \].

Key concepts

Latent Heat of Fusion

Latent heat of fusion is the amount of energy required to change a substance from a solid to a liquid or vice versa without changing its temperature. In the context of ice formation, it refers to the energy needed to freeze water into ice.

This energy is absorbed during the phase change process, and in this exercise, it is crucial as it determines how much heat has to be removed to form a specific thickness of ice. The formula for latent heat of fusion is usually expressed as:

• \[ Q = mL \]

where:

• \( Q \) is the heat absorbed or released,
• \( m \) is the mass,
• \( L \) is the latent heat of fusion.

In this exercise, we often express this concept in terms of "mass per unit area" as \( m = \rho \cdot A \cdot d \), where \( \rho \) is the density, \( A \) is the area, and \( d \) is the ice thickness. This helps relate heat loss to the formation of an ice layer.

Thermal Conductivity

Thermal conductivity is a material property that indicates its ability to conduct heat. It is denoted by \( k \), and it represents the amount of heat (in joules) that passes through a material per unit time, thickness, and temperature difference.

In general terms, a material with high thermal conductivity conducts heat well, while a low value means the material is a good insulator. The equation used is:

• \[ Q = \frac{kA(T_s - T_i)t}{d} \]

where:

• \( Q \) is the heat transferred,
• \( A \) is the cross-sectional area of heat flow,
• \( T_s \) and \( T_i \) are surface and initial temperatures respectively,
• \( t \) is time,
• \( d \) is the thickness of the material.

This means the rate at which ice forms depends on the thermal conductivity of the ice, which is part of why we use this property to calculate ice formation.

Ice Thickness Calculation

Calculating ice thickness involves solving for \( d \) (thickness of the ice layer) in the exercise. The formula derived combines concepts of thermal conductivity and latent heat to equate energy expressions.

From the equation \( Q = \frac{kA(T_s - T_i)t}{d} = \rho A d L \), after canceling \( A \), the ice thickness formula is deduced:

• \[ d^2 = \frac{k(T_s - T_i)t}{\rho L} \]

where:

• \( k \) is the thermal conductivity,
• \( T_s \) is the surface temperature,
• \( T_i \) is the initial temperature,
• \( t \) is the time,
• \( \rho \) is the density,
• \( L \) is the latent heat of fusion.

Once \( d^2 \) is found, taking the square root provides \( d \), the ice thickness.

Temperature Gradient in Solids

The temperature gradient in solids is defined as the rate of change of temperature with respect to distance within a solid material. It's represented mathematically by:

• \[ \text{temperature gradient} = \frac{\Delta T}{\Delta x} \]

where:

• \( \Delta T \) is the temperature difference,
• \( \Delta x \) is the thickness or distance across the material.

In the context of ice formation, the temperature gradient drives the heat transfer through the ice. This is because heat naturally moves from a warmer area to a cooler one, which, in this case, is from the water at \(0^{\circ} \mathrm{C}\) to the surface at \(-10^{\circ} \mathrm{C}\).

Understanding the gradient helps in predicting how quickly energy is transferred, thus how rapidly ice forms. It links directly to the thermal conductivity and the calculated ice thickness, showing the interplay of these factors in a practical scenario.