Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.
Does Thermal Conductivity Change With Thickness?
Thickness of a slab can vary during thermal conduction.
In cold countries when temperature falls below 0°C, the water on the surface of the lakes and ponds slowly freeze to ice. Heat from rest of the mass of water is conducted to the atmosphere through the top frozen layer and the thickness of the ice layer on the surface gradually increases. So this is a fine example of conduction of heat through a slab of varying thickness. When the temperature of atmosphere falls below 0°C, the uppermost surface of water loses its latent heat to the atmosphere and forms a thin sheet of ice. In this way, as the thickness of the layer of ice increases, heat will have to be conducted through a thicker layer too.
Let the coefficient of conductivity of ice = k, density of ice at 0°C = ρ, the temperature of air above the ice surface = -θ°C and uppermost surface area of the water body = A [Fig.].
After a time t s from the beginning of formation of ice, let the thickness of ice formed on the water body be x.
If in a small interval of time dt, the increase in thickness of ice is dx, then the volume of ice formed in dt is Adx.
∴ The mass of that ice = Aρdx.
Therefore, the heat lost by water or the heat conducted to air in time dt through ice of thickness x is,
dQ = Aρdx ᐧ L [where L = latent heat of ice]
∴ Rate of flow of heat,
\(\frac{d Q}{d t}\) = AρL\(\frac{d x}{d t}\) = \(\frac{k A[0-(-\theta)]}{x}\) = \(\frac{k A \theta}{x}\)
or, xdx = \(\frac{k}{\rho L}\)θdt ……. (1)
i) Let us assume that at t = 0, x = x1 and at t = t1, x = x2. This means that in t1 s time if the thickness of ice slab has increased from x1 to x2, integrating the equation (1) we get,
\(\int_{x_1}^{x_2} x d x\) = \(\int_0^{t_1} \frac{k \theta}{\rho L} d t\)
or, \(\frac{1}{2}\left(x_2^2-x_1^2\right)\) = \(\frac{k \theta}{\rho L} t_1\)
∴ t1 = \(\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right)\) …….. (2)
ii) If x1 = 0 when t = 0 and x2 = x when t = t1
t1 = \(\frac{\rho L}{2 k \theta} x^2\) ……….. (3)
This equation gives the time (t1) required for deposition of a layer of ice of thickness x.
Note that in both equations (1) and (2), t1 is independent of the surface area A of the lake. So in the same weather condition, water bodies of all sizes from ice in this same rate.
Numerical Examples
Example 1.
A 3 cm thick layer of ice is formed over a water reservoir. Temperature over the reservoir is -20°C. In what time, thickness of the ice layer will increase by 1 mm? Conductivity of ice = 0.005 CGS unit; latent heat of fusion of ice = 80 cal ᐧ g-1; density of ice at 0°C = 0.91 g ᐧ cm-3.
Solution:
Example 2.
The surface of a lake is covered with a layer of ice of thickness 10 cm. For increase in thickness of the ice layer by 1 mm, time taken is found to be 49 min 9 s. Conductivity of ice = 0.005 CGS unit, latent heat of fusion of ice = 80 cal ᐧ g-1 and density of ice = 0.9 17 g ᐧ cm-3. Find the outside temperature.
Solution:
Let temperature of air outside = -θ
It is known that,
In this case, t = 49 min 9 s = 2949 s, x2 = 10.1cm, x1 = 10 cm, ρ = 0.917 g ᐧ cm-3, L = 80 cal ᐧ g-1, k = 0.005 CGS unit
∴ θ = \(\frac{0.917 \times 80}{2 \times 0.005 \times 2949}\)[(10.1)2 – (10)2]
= \(\frac{0.917 \times 80 \times 2.01}{2 \times 0.005 \times 2949}\) = 5
∴ Outside temperature is -5°C.
Example 3.
Water in a tank at 0°C is in contact with a surrounding temperature of -20°C. Prove that the rate of increase of thickness x (in cm) of ice on the surface, is related to time t (in s) as x2 = 0.00273 t. Density of ice = 0.917 g ᐧ cm-3, latent heat of fusion of ice = 80 cal ᐧ g-1 and conductivity of ice = 0.005 CGS unit.
Solution:
Let, A be the surface area of the tank. x be the initial thickness of ice floating on the tank surface and dx be the increase in thickness in time dt.
Hence, mass of ice formed in the time dt = Adx × ρ [ρ = density of ice]
Heat released by water Adx × ρ × L [L = latent heat of fusion of ice]
This is the amount of heat that is conducted from water in the tank through the ice of thickness x to the surroundings.
∴ Rate of release of heat,
Example 4.
A 14.9 cm thick layer of ice floats on a deep lake. The temperature of the upper surface of the ice layer is the same as that of the surrounding air. If this temperature remains constant at -1 °C then determine the time required for the layer to increase by 2 mm in thickness. Given, latent heat of melting of ice = 80 cal ᐧ g-1, density of ice = 0.9 g ᐧ cm-3 and coefficient of thermal conductivity of ice = 0.006 CGS unit.
Solution:
We know, t = \(\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right)\)
Here, ρ = 0.9 g ᐧ cm-3, L = 80 cal ᐧ g-1
k = 0.006 CGS unit, θ = 1°C
x1 = 14.9 cm, x2 = 14.9 + 0.2 = 15.1 cm
∴ t = \(\frac{0.9 \times 80}{2 \times 0.006 \times 1}\)[(15.1)2 – (14.9)2]
= \(\frac{0.9 \times 80}{0.012}\) × 30 × 0.2 = 36000 s = 10 h