Question

On a winter day when the atmospheric temperature drops to -10°C, ice forms on the surface of a lake.

(a) Calculate the rate of increase of thickness of the ice when 10 cm of ice is already formed. (b) Calculate the total time taken in forming 10 cm of ice. Assume that the temperature of the entire water reaches 0°C before the ice starts forming. Density of water = 1000 kg m$^{-3}$, latent heat of fusion of ice = 3.36×10$^5$ J kg$^{-1}$ and thermal conductivity of ice = 1.7 W m$^{-1}$°C$^{-1}$. Neglect the expansion of water on freezing.

Consider the situation of the previous problem. Assume that the temperature of the water at the bottom of the lake remains constant at 4°C as the ice forms on the surface (the heat required to maintain the temperature of the bottom layer may come from the bed of the lake). The depth of the lake is 1.0 m. Show that the thickness of the ice formed attains a steady state maximum value. Find this value. The thermal conductivity of water = 0.50 W m$^{-1}$°C$^{-1}$. Take other relevant data from the previous problem.

Answer 1

(a) The rate of increase of thickness of the ice when 10 cm of ice is already formed is approximately $5.06 \times 10^{-7} \text{ m/s}$. (b) The total time taken in forming 10 cm of ice is approximately $27.45 \text{ hours}$. (c) The thickness of the ice formed attains a steady state maximum value. This value is approximately $0.8947 \text{ m}$.

Answer 2

The problem consists of two distinct parts.

Part 1: Ice formation with water at 0°C

(a) Rate of increase of thickness of ice: Let $x$ be the thickness of the ice layer at time $t$. The temperature difference across the ice layer is $\Delta T_{ice} = 0^\circ C - (-10^\circ C) = 10^\circ C$. The rate of heat conduction through the ice layer of area $A$ is given by: $\frac{dQ}{dt} = \frac{k_{ice} A \Delta T_{ice}}{x}$ This heat loss causes water to freeze at the ice-water interface. If an additional thickness $dx$ of ice forms in time $dt$, the mass of ice formed is $dm = \rho A dx$. The heat released during this freezing is $dQ = dm L_f = \rho A dx L_f$. So, the rate of heat released by freezing is: $\frac{dQ}{dt} = \rho A L_f \frac{dx}{dt}$ Equating the two rates: $\frac{k_{ice} A \Delta T_{ice}}{x} = \rho A L_f \frac{dx}{dt}$ $\frac{dx}{dt} = \frac{k_{ice} \Delta T_{ice}}{\rho L_f x}$ Given: $k_{ice} = 1.7 \text{ W m}^{-1}°C^{-1}$, $\Delta T_{ice} = 10^\circ C$, $\rho = 1000 \text{ kg m}^{-3}$, $L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$. When $x = 10 \text{ cm} = 0.1 \text{ m}$: $\frac{dx}{dt} = \frac{1.7 \times 10}{1000 \times 3.36 \times 10^5 \times 0.1} = \frac{17}{3.36 \times 10^7} \approx 5.06 \times 10^{-7} \text{ m/s}$

(b) Total time taken in forming 10 cm of ice: From the rate equation: $x \ dx = \frac{k_{ice} \Delta T_{ice}}{\rho L_f} \ dt$ Integrate from $x=0$ to $x=X$ and $t=0$ to $t=T$: $\int_0^X x \ dx = \frac{k_{ice} \Delta T_{ice}}{\rho L_f} \int_0^T dt$ $\frac{X^2}{2} = \frac{k_{ice} \Delta T_{ice}}{\rho L_f} T$ $T = \frac{X^2 \rho L_f}{2 k_{ice} \Delta T_{ice}}$ For $X = 10 \text{ cm} = 0.1 \text{ m}$: $T = \frac{(0.1)^2 \times 1000 \times 3.36 \times 10^5}{2 \times 1.7 \times 10} = \frac{0.01 \times 1000 \times 3.36 \times 10^5}{34}$ $T = \frac{10 \times 3.36 \times 10^5}{34} = \frac{3.36 \times 10^6}{34} \approx 98823.5 \text{ s}$ Converting to hours: $T = \frac{98823.5}{3600} \approx 27.45 \text{ hours}$

Part 2: Steady state maximum thickness of ice

Let $x$ be the thickness of the ice layer and $(D-x)$ be the thickness of the water layer, where $D = 1.0 \text{ m}$ is the total depth of the lake. At steady state, the temperature at the ice-water interface is $0^\circ C$. The rate of heat flow from the bottom of the lake ($T_{bottom} = 4^\circ C$) to the ice-water interface ($0^\circ C$) through the water layer is: $\left(\frac{dQ}{dt}\right)_{water} = \frac{k_{water} A (T_{bottom} - 0)}{D-x} = \frac{k_{water} A (4)}{D-x}$ The rate of heat flow from the ice-water interface ($0^\circ C$) to the atmosphere ($T_a = -10^\circ C$) through the ice layer is: $\left(\frac{dQ}{dt}\right)_{ice} = \frac{k_{ice} A (0 - T_a)}{x} = \frac{k_{ice} A (10)}{x}$ For a steady state, these two rates must be equal: $\frac{4 k_{water} A}{D-x} = \frac{10 k_{ice} A}{x}$ $\frac{4 k_{water}}{D-x} = \frac{10 k_{ice}}{x}$ $4 k_{water} x = 10 k_{ice} (D-x)$ $4 k_{water} x = 10 k_{ice} D - 10 k_{ice} x$ $x (4 k_{water} + 10 k_{ice}) = 10 k_{ice} D$ $x = \frac{10 k_{ice} D}{4 k_{water} + 10 k_{ice}}$ This equation shows that the thickness $x$ is a constant value determined by the given parameters ($k_{ice}, k_{water}, D$). Therefore, the thickness of the ice formed attains a steady state value. To show it's a maximum value: If $x$ were less than this steady-state value, the rate of heat conduction through the ice would be greater than the rate of heat supply from the water, leading to further freezing. If $x$ were greater, the rate of heat supply from the water would exceed the rate of heat conduction through the ice, causing the ice to melt from the bottom. Thus, this steady state value is the maximum stable thickness.

Calculate the maximum thickness: Given: $k_{ice} = 1.7 \text{ W m}^{-1}°C^{-1}$, $k_{water} = 0.50 \text{ W m}^{-1}°C^{-1}$, $D = 1.0 \text{ m}$. $x = \frac{10 \times 1.7 \times 1.0}{4 \times 0.50 + 10 \times 1.7} = \frac{17}{2.0 + 17.0} = \frac{17}{19}$ $x \approx 0.8947 \text{ m}$

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Explanation of the solution:

1. Part 1 (a) Rate of increase of ice thickness: Equate the rate of heat conducted through the ice layer to the rate of latent heat released by freezing water. The heat conduction rate is $Q/t = k_{ice} A \Delta T / x$. The heat released by freezing is $Q = \rho A dx L_f$. Equating these gives a differential equation for $dx/dt$. Substitute given values for $x = 0.1 \text{ m}$.
2. Part 1 (b) Total time to form 10 cm of ice: Integrate the differential equation obtained in part (a) from $x=0$ to $x=0.1 \text{ m}$ to find the total time $T$.
3. Part 2 (Steady state thickness): At steady state, no more ice forms, implying the ice-water interface is at $0^\circ C$. The rate of heat flow from the bottom water (at $4^\circ C$) through the water layer to the interface must equal the rate of heat flow from the interface through the ice layer to the atmosphere (at $-10^\circ C$). Set these two heat flow rates equal to each other to solve for the steady-state ice thickness $x$. The resulting equation for $x$ depends only on constants, showing a steady state is reached. The argument for it being a maximum involves considering what happens if $x$ deviates from this equilibrium value.