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## Homework Statement

ice needs to be at least 6.76 feet thick. = 2.0604m

Currently the ice is 5.78 feet thick = 1.7617m

Later, there is a cold spell

If the average temperature during the cold spell is 23.23°F =-4.8722C = 268.277K

and the water temperature is 32.0°F = 0C = 273.15K

how many hours will it take for the frozen lake to be safely passable?

The latent heat of fusion of water is 3.33 × 105 J/kg

and the thermal conductivity of ice is 2.180 W/m·K.

The density of water near freezing is 1000.0 kg/m3 and the

density of ice is 917.00 kg/m3.

This problem involves calculus. Let y denote the thickness of the ice at any time t, with dy the thickness of an infinitesimal layer of ice that resides just above the water surface. Each layer freezes through the release of heat from the water to the air above. Therefore, the heat needed to be released by the water to freeze the infinitesimal layer dy must equal the heat that passes through the layer of thickness y above it.

## Homework Equations

celsius > Kelvin = +273.15

heat released via conduction = k * Area * (Thot - Tcold) / L

k= thermal conductivity

L = meters

## The Attempt at a Solution

In kelvin , T 1 = 23.23°F = 268.277 K

T2 = 32.0°F = 273.15 K

heat released to air = k * A * (T1- T2) / L = 2.180 * A * ( 273.15 - 268.277) /L = 10.623*A / L Watts.

we have dQ = -(10.623 / (L ^2)) dL .

heat lost = heat utilized for freezing

L is length of the ice ,

so, if Q = heat lost = (dm/dt)* latent heat of fusion of water = ( A* dL * density /dt ) * latent heat of fusion of water

Q = ( 1000*A* dL/dt) * 3.33 × 10^5

so, 10.623*A / L = ( 1000*A* dL/dt) * 3.33 × 10^5

dL/dt = (10.623/(1000* 3.33 × 10^5))/L

so, dL/L = 0.0000000319 * dt

integrating ln L = 0.0000000319 * t + constant.

At time t =0 , L = 5.78 feet = 1.7617 m

ln 1.7617 = 0.0000000319 * 0 + constant

constant = ln 1.7617 = 0.566279 .

so, ln L = 0.0000000319 * t + 0.566279

at time t, L = 6.76 feet = 2.06 m

so, ln 2.06 = 0.0000000319 * t + 0.566279

time , t = 4903528.69 seconds = 1362.09 hours = 56.7538 days .

It takes 1362.09 hours for the ice to become 6.76 feet thick and be safe for driving.

this seems like alot of time just to get another foot of ice on the lake when the temp of the air i 9 degrees F below freezing.