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

See the solution for detailed answer

Answer 2

The problem involves two scenarios of ice formation on a lake.

Problem 1: Ice formation with constant atmospheric temperature

Concept: Heat flows from the water (at $0^\circ C$) through the ice layer to the colder atmosphere (at $-10^\circ C$). This heat removal causes water to freeze at the ice-water interface.

Let $x$ be the thickness of the ice layer at time $t$. The rate of heat flow through the ice layer of area $A$ is given by Fourier's law: $\frac{dQ}{dt} = \frac{k_i A (T_{water} - T_{atmosphere})}{x}$ where $k_i$ is the thermal conductivity of ice, $T_{water} = 0^\circ C$, and $T_{atmosphere} = -10^\circ C$. So, $T_{water} - T_{atmosphere} = 10^\circ C$.

The heat removed is used to freeze water. If the ice thickness increases by $dx$ in time $dt$, the mass of water frozen is $dm = \rho_i A dx$, where $\rho_i$ is the density of ice (assumed equal to water's density, $\rho_w$). The heat required for this freezing is $dQ = dm \cdot L_f = \rho_i A dx L_f$. So, the rate of heat removal for freezing is: $\frac{dQ}{dt} = \rho_i A L_f \frac{dx}{dt}$ Equating the two expressions for $\frac{dQ}{dt}$: $\frac{k_i A (T_{water} - T_{atmosphere})}{x} = \rho_i A L_f \frac{dx}{dt}$ $\frac{dx}{dt} = \frac{k_i (T_{water} - T_{atmosphere})}{\rho_i L_f x}$

Given Data: $T_{atmosphere} = -10^\circ C$ $T_{water} = 0^\circ C$ $\rho_i = \rho_w = 1000 \text{ kg m}^{-3}$ $L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$ k_i = 1.7 \text{ W m}^{-1} ^\circ C^{-1}

(a) Calculate the rate of increase of thickness of the ice when 10 cm of ice is already formed. Here, $x = 10 \text{ cm} = 0.1 \text{ m}$. \frac{dx}{dt} = \frac{1.7 \text{ W m}^{-1} ^\circ C^{-1} \times (0 - (-10))^\circ C}{1000 \text{ kg m}^{-3} \times 3.36 \times 10^5 \text{ J kg}^{-1} \times 0.1 \text{ m}} $\frac{dx}{dt} = \frac{1.7 \times 10}{1000 \times 3.36 \times 10^5 \times 0.1} = \frac{17}{3.36 \times 10^7}$ $\frac{dx}{dt} \approx 5.06 \times 10^{-7} \text{ m/s}$ To convert to cm/hour: $5.06 \times 10^{-7} \text{ m/s} \times \frac{100 \text{ cm}}{1 \text{ m}} \times \frac{3600 \text{ s}}{1 \text{ hour}} \approx 0.182 \text{ cm/hour}$

(b) Calculate the total time taken in forming 10 cm of ice. From the differential equation: $x \ dx = \frac{k_i (T_{water} - T_{atmosphere})}{\rho_i L_f} \ dt$ Integrate both sides from $x=0$ to $x=0.1 \text{ m}$ and from $t=0$ to $t=t_f$: $\int_0^{0.1} x \ dx = \frac{k_i (T_{water} - T_{atmosphere})}{\rho_i L_f} \int_0^{t_f} dt$ $\left[ \frac{x^2}{2} \right]_0^{0.1} = \frac{k_i (T_{water} - T_{atmosphere})}{\rho_i L_f} t_f$ $\frac{(0.1)^2}{2} = \frac{1.7 \times 10}{1000 \times 3.36 \times 10^5} t_f$ $\frac{0.01}{2} = \frac{17}{3.36 \times 10^8} t_f$ $0.005 = \frac{17}{3.36 \times 10^8} t_f$ $t_f = \frac{0.005 \times 3.36 \times 10^8}{17} = \frac{1.68 \times 10^6}{17}$ $t_f \approx 98823.5 \text{ seconds}$ To convert to hours: $t_f = \frac{98823.5}{3600} \approx 27.45 \text{ hours}$

Problem 2: Steady state maximum thickness of ice

Concept: In this scenario, heat flows from the bottom of the lake (at $4^\circ C$) through the water layer to the ice-water interface (at $0^\circ C$). Simultaneously, heat flows from the ice-water interface through the ice layer to the atmosphere (at $-10^\circ C$). A steady state is reached when the rate of heat flow from the bottom of the lake to the ice-water interface equals the rate of heat flow from the ice-water interface to the atmosphere. At this point, no net heat is available for freezing, and the ice thickness reaches a maximum constant value.

Let $x_{max}$ be the maximum thickness of the ice. The depth of the lake is $D = 1.0 \text{ m}$. The thickness of the water layer below the ice is $(D - x_{max})$.

Rate of heat flow through water ($Q_w/t$): $T_{bottom} = 4^\circ C$, $T_{ice-water} = 0^\circ C$. $\frac{Q_w}{t} = \frac{k_w A (T_{bottom} - T_{ice-water})}{D - x_{max}} = \frac{k_w A (4^\circ C - 0^\circ C)}{D - x_{max}} = \frac{k_w A \cdot 4}{D - x_{max}}$ Rate of heat flow through ice ($Q_i/t$): $T_{ice-water} = 0^\circ C$, $T_{atmosphere} = -10^\circ C$. $\frac{Q_i}{t} = \frac{k_i A (T_{ice-water} - T_{atmosphere})}{x_{max}} = \frac{k_i A (0^\circ C - (-10^\circ C))}{x_{max}} = \frac{k_i A \cdot 10}{x_{max}}$ At steady state, $\frac{Q_w}{t} = \frac{Q_i}{t}$: $\frac{k_w A \cdot 4}{D - x_{max}} = \frac{k_i A \cdot 10}{x_{max}}$ $\frac{k_w \cdot 4}{D - x_{max}} = \frac{k_i \cdot 10}{x_{max}}$ $4 k_w x_{max} = 10 k_i (D - x_{max})$ $4 k_w x_{max} = 10 k_i D - 10 k_i x_{max}$ $(4 k_w + 10 k_i) x_{max} = 10 k_i D$ $x_{max} = \frac{10 k_i D}{4 k_w + 10 k_i}$

Given Data (additional for this problem): k_w = 0.50 \text{ W m}^{-1} ^\circ C^{-1} $D = 1.0 \text{ m}$ (Other data k_i = 1.7 \text{ W m}^{-1} ^\circ C^{-1} from previous problem)

Show that the thickness of the ice formed attains a steady state maximum value: As the ice thickness $x$ increases, the rate of heat loss to the atmosphere ($Q_i/t \propto 1/x$) decreases. Simultaneously, the rate of heat supply from the lake bottom ($Q_w/t \propto 1/(D-x)$) increases (because $D-x$ decreases). When the rate of heat lost to the atmosphere becomes equal to the rate of heat supplied from the lake bottom, there is no net heat available for freezing, and thus, the ice growth stops, leading to a steady state maximum thickness.

Find this value: x_{max} = \frac{10 \times 1.7 \text{ W m}^{-1} ^\circ C^{-1} \times 1.0 \text{ m}}{4 \times 0.50 \text{ W m}^{-1} ^\circ C^{-1} + 10 \times 1.7 \text{ W m}^{-1} ^\circ C^{-1}} $x_{max} = \frac{17}{2.0 + 17.0} = \frac{17}{19}$ $x_{max} \approx 0.8947 \text{ m} \approx 89.5 \text{ cm}$

Explanation of the solution:

Problem 1: (a) The rate of ice formation ($dx/dt$) is determined by equating the heat conducted through the existing ice layer to the heat required to freeze a new layer of water. The formula derived is $dx/dt = \frac{k_i (T_{water} - T_{atmosphere})}{\rho_i L_f x}$. Substituting the given values for $x=0.1$ m yields the rate. (b) To find the total time, the differential equation for $dx/dt$ is integrated from $x=0$ to $x=0.1$ m. The result is $t_f = \frac{x_f^2 \rho_i L_f}{2 k_i (T_{water} - T_{atmosphere})}$.

Problem 2: The ice thickness attains a steady state when the heat flowing from the warmer lake bottom through the water layer to the ice-water interface exactly balances the heat flowing from the ice-water interface through the ice layer to the colder atmosphere. At this point, no more water freezes. The heat flow equations for conduction through water and ice are set equal, and the resulting algebraic equation is solved for $x_{max}$. As $x$ increases, heat loss through ice decreases, and heat gain from water increases, leading to a balance point.

Answer:

Problem 1: (a) Rate of increase of thickness of ice when 10 cm of ice is already formed: $0.182 \text{ cm/hour}$ (b) Total time taken in forming 10 cm of ice: $27.45 \text{ hours}$

Problem 2: The thickness of the ice formed attains a steady state maximum value because as ice thickness increases, the rate of heat loss to the atmosphere decreases, while the rate of heat supply from the lake bottom increases. Eventually, these two rates become equal, stopping further ice growth. The steady state maximum value of ice thickness is: $0.895 \text{ m}$ or $89.5 \text{ cm}$