The water at the surface of a lake and the air above it are...

The water at the surface of a lake and the air above it are in thermal equilibrium just above the freezing point. The air temperature suddenly drops by $\Delta T$ degrees. Find the thickness of the ice on the lake as a function of time in terms of the latent heat per unit volume $L / V$ and the thermal conductivity $\Lambda$ of the ice. Assume that $\Delta T$ is small enough that the specific heat of the ice may be neglected.

Key Concepts

Fourier’s Law of Heat Conduction

Fourier’s Law of heat conduction states that the heat flux within a material is proportional to the negative temperature gradient. This principle provides the mathematical framework to model heat transfer through the ice, linking the temperature difference to the rate of energy conduction necessary for phase changes.

Thermal Conductivity

Thermal conductivity is an intrinsic material property that indicates how well a material conducts heat. It quantifies the rate at which heat is transferred through a material when subjected to a temperature gradient, making it fundamental in problems involving heat conduction through solids like ice.

Latent Heat

Latent heat is the amount of energy per unit volume or mass that is required for a substance to undergo a phase change at a constant temperature. It plays a crucial role in phase-change phenomena by accounting for the energy released or absorbed during the transition, independent of temperature changes in the material.

Stefan Problem

The Stefan problem is a class of free boundary problems that deals with phase-change scenarios, such as the freezing and melting of a liquid. It involves determining both the temperature distribution and the moving interface between phases by applying energy conservation principles at the phase boundary.

Transcript

00:01 Hi everyone this is the problem based on change of phase of a substance due to transfer of heat here it is given the water in the lake gets freeze when heat flows from water to air so first part the entire lake does not freeze because the heat transferred by conduction between cold air and top layer of water so it is just upper layer of water fridges to be part to prove thickness of ice deposited on water surface is directly opposed to square root temperature let us here if we assume delta m dm mass of water gets freezed then mass of water freeze is defined as density into volume density is row a into d l d l is the small thickness of water which gets freezed you can see here so delta is this thickness of layer which gets freezed so amount of the heat required to freeze will be dm into latent heat of fusion so it will be row a latent heat of fusion into d l and amount of the heat transferred per unit time will be k a delta t upon l substitute the value d q row a latent heat of fusion d l upon d t upon d t is called to k a delta t upon l cross multiply row a latent heat of fusion l d l to be k a data t into integrating it for the limit 0 to t and 0 to l here it is a let me correct so you will get row a latent heat of fusion into l square to be equal to k a delta t into t so l square will be twice of k a delta t this is the changing temperature upon row into a latent heat of fuzzle this is constant so l will be l square is directly proportional to t so l will be directly proportional to square root of t.

05:06 Now here it is given temperature of ice is this is given in the problem minus 10 degrees celsius temperature of air is 0 degrees celsius thickness is 0 .25 so time required will be n square into density into latent heat of fusion upon twice of k into data t substitute the value length is given 0 .25 square density is 9 to 0 density of ice you have to take latent heat of fusion 3 34 into 10 to the power 3 divided by conductive of the coefficient of conduction of ice that is 1 .6 temperature will be 0 minus 10 so it is to be equal to 600 into 10 to the power 3 second or it can be written as 166 .6 hour depart if depth of the lake is 40 meter then time required to freeze.

06:48 Freeze completely using the relation and square row...

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