How many hours until ice on lake freezes from 5.78ft to 6.76ft?

[Sorbik]

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 a lot of time just to get another foot of ice on the lake when the temp of the air i 9 degrees F below freezing.

 

[lightgrav]

Ice is a poor conductor (I think its k=1.6 W/Km), but 2 meters is really thick insulation.
You can check your calculus by using a "back-of-envelope" algebra solution;
... the extra ¼ meter is small relative to the average thickness ... .
I would use y for the thickness (reserving L for Latent heat): dtime = Q/Power .
your conduction [ k * A * (T1- T2) / L = ... Watts] is Power, not Energy Q.

With your value for k, I get approximately 15 million seconds.

 

[haruspex]

You make it very hard to follow by plugging in constants right from the start. Algebra is much clearer if you work entirely symbolically, only plugging in values at the end.

heat released to air = k * A * (T1- T2) / L, where L is length of ice

I guess you mean rate of release of heat (dQ/dt), L is depth of ice, and A is some arbitrary area.

we have dQ = -(k * A * (T1- T2) / (L ^2)) dL .

How do you get that?

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

I suspect the answer is even greater.

 

[Sorbik]

I'll rework this and try to keep symbols to the end.

 

[Sorbik]

My calc understing is pretty poor. Lots of bad turns. I don't think this problem should have a ln(L) in it either after integrating or have the need to use indefinite integrals and solving for constants.

 

[haruspex]

You're still messing with dQ/dL (or maybe it's (d/dL)(dQ/dt)). That does not mean anything useful.
You have, correctly, $\dot Q = k A \Delta\theta / L$ and $\dot Q = \rho A \dot L \Lambda_f$ (though you wrote Q instead of $\dot Q$ in each case), where Λ_f is the latent heat of fusion, and ρ is the density. Combine those two equations and solve the differential equation in L and t. You should not get any logarithmic terms.

 

[lightgrav]

your time is in the wrong place, because you confused Power conduction with Energy,
and you confused Energy for freezing with rate (you wrote dm/dt, instead of dm).

 

[cjl]

Ice takes a long time to form to a large thickness, especially if the ambient temperature isn't far below freezing. That having been said, ice doesn't need to be anywhere near 2 meters thick to be safe for driving. If I remember right, the actual requirement is more like ten or twelve inches of ice (but I could be off a bit, since I'm going by memory here).

 

[Chestermiller]

It looks like you pretty much knew what you were doing. You just made an algebraic error between the following two lines:

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

so, dL/L = 0.0000000319 * dt

It should be LdL, not dL/L.

Otherwise, it looks correct.

Chet

 

[Sorbik]

Ok.

I have LdL = 3.19x10^-8*dt

∫LdL ==> L^2 / 2 = 3.19x10^-8 *t + C

at time = 0, L = 1.7617m

1.7617^2 / 2 = 0 + C ===> 1.55179 = C

at time = t, L = 2.0604m

2.0604^2 / 2 = 3.19x10^-8t + 1.55179

t = 17,894,377s ==> 4970.66 hours <-- time it takes to get another foot of ice ... 207.111 days?

 

[Chestermiller]

Ok.

I have LdL = 3.19x10^-8*dt

∫LdL ==> L^2 / 2 = 3.19x10^-8 *t + C

at time = 0, L = 1.7617m

1.7617^2 / 2 = 0 + C ===> 1.55179 = C

at time = t, L = 2.0604m

2.0604^2 / 2 = 3.19x10^-8t + 1.55179

t = 17,894,377s ==> 4970.66 hours <-- time it takes to get another foot of ice ... 207.111 days?

Looks correct. That's a lot longer than winter, though. All that means is that the ice at this location will never get that thick (under present climate conditions).

Chet

 

[Sorbik]

thanks for the help. got it to work

 

[D H]

Ok.

I have LdL = 3.19x10^-8*dt

That's close. You are not accounting for the fact that a cubic meter of water becomes more than a cubic meter of ice.

t = 17,894,377s ==> 4970.66 hours <-- time it takes to get another foot of ice ... 207.111 days?

That's an extremely long time. The problem statement is not very realistic.

One issue is that 23.23 Fahrenheit does not qualify as a "cold snap" in regions where people use ice roads. That's a hot winter day in the far north. A cold snap would be -40 F (or -40 C, same thing) or colder.

Another issue is that ice roads don't need to be nearly two meters thick. For example, the Tibbitt to Contwoyto Winter Road needs to be a bit over a meter thick to handle very, very large trucks. From http://www.jvtcwinterroad.ca/jvwr/ ,

Based on the minimum ice thickness along the entire route, acceptable load weight limits are set. With each additional inch of ice, the allowable weight increases. When a thickness of 70 cms (27-28 inches) is achieved over the entire road, very light loads known as ‘hotshots’ are dispatched. When the ice reaches 107 cms (42 inches) along the entire road, it is thick enough for a super B tanker fully loaded with 48,000 to 50,000 litres of fuel. A super B is a tractor hauling two tanks of fuel weighs approximately 41-42 tonnes.
​

That said, an ice road designed to handle heavy trucks such as the cited road are only open for a couple of months every year because it does indeed take a long time to form even a meter thick layer of ice.