When the air temperature is below 0 C, the water at the surface of a lake freezes to form a sheet of ice.
If the upper surface of an ice sheet 24.2 cm thick is at -10.7 C and the bottom surface is at 0.00 C, calculate the time it will take to add 1.80 mm to the thickness of this sheet.
(2.42cm=.0242m)
(1.80mm=.00180m)
I don't have any equations with time in my notes, any ideas????
if anyone checks this-please help!!!! please!!
bnbn
To calculate the time it will take to add 1.80 mm to the thickness of the ice sheet, we can use the concept of heat conduction. Heat conduction is the transfer of heat from one object to another through direct contact. In this case, the heat will be transferred from the air to the ice sheet, causing it to melt and increase in thickness.
To solve this problem, we can use the equation for heat conduction:
Q = (k * A * ΔT * t) / d
where Q is the amount of heat transferred, k is the thermal conductivity of the material (in this case, ice), A is the surface area through which heat flows, ΔT is the temperature difference between the ice sheet and the air, t is the time, and d is the thickness of the ice sheet.
We need to solve for t, so we can rearrange the equation as follows:
t = (Q * d) / (k * A * ΔT)
Now let's substitute the given values into the equation:
Q = heat transferred = 1.80 mm of ice sheet thickness (converted to meters) = 0.00180 m
d = initial ice sheet thickness = 24.2 cm (converted to meters) = 0.242 m
k = thermal conductivity of ice = 2.2 W/(m·K) (this value can be looked up or approximated)
A = surface area = assumed to be 1 m² (since the equation does not explicitly specify it)
ΔT = temperature difference = -10.7°C - 0.0°C = -10.7°C (converted to Kelvin) = 262.45 K
Now substitute the values into the equation:
t = (0.00180 * 0.242) / (2.2 * 1 * 262.45)
t ≈ 0.000195 seconds
Therefore, it will take approximately 0.000195 seconds to add 1.80 mm to the thickness of the ice sheet.
Assume that the freezing of an additional layer is slow enough. Then the temperature profile can be assumed to be linear across the layer of ice. This quasistatic assumptions allows us to solve this problem without the need of solving the much more difficult moving boundary condition.
The amount of heat that leaves the top surface has two origins: the latent heat of freezing of the additional sheet of water at the bottom plus the heat from additional cooling of the existing ice.
This will give a heat balance equation from which a "thickness goes as sqrt(time) " formula emerges.
First, the linear temperature profile assumed is (meauring thickness x from the top surface downwards)
T = T0 + x delta(T)/d
The energy balance is for infinitesimal time delta(t) is
k (delta(T)/d) delta(t) = rho L delta(d) + integral_0^d rho sigma delta(T) dx
d is the thickness of the sheet (distance of growth)
With the linear profile this gives
d(t) ^2 = d(0)^2 + t * 2 k delta(T)/( rho L + 1/2 rho sigma delta(T) )
Here k is the conductivity of ice, L the latent heat of melting, rho the density of ice and delta(T) the temperature difference between top and bottom.
From this you can calcute the time it takes to grow an additional 2.10 mm on 24.6 cm.
Just solve for t with d(0) = 0.246 m and d(t) = 0.2481 m