Icicles.Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis (Fig. 18-55). Because the water–ice interface must have a temperature of 0 C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions. It can lose energy and freeze

only by sending energy up (through distance L) to the top of the icicle, where the temperature ${\mathit{T}}_{\mathbf{r}}$can be below ${\mathbf{\text{0}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$.Take$\mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{}\mathbf{\text{m}}$and${\mathit{T}}_{\mathbf{r}}\mathbf{=}{\mathbf{5}}^{\mathbf{o}}\mathit{C}$ . Assume that the central tube and the upward conduction path both have cross-sectional area A. In terms of A, what rate is (a) energy conducted upward and (b) massconvertedfrom liquid to ice at the top of the central tube? (c) At what rate does the top of the tube move downward because of water freezing the central axis (Fig. 18-55). Because the water–ice interface must have a temperature of ${\mathbf{0}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$, the water in the tube cannot lose energy through there? The thermal conductivity of ice is $\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{ }\mathbf{}\mathbf{\text{W/m}}\mathbf{=}\mathit{K}$, and the density of liquid water is $\mathbf{1000}\mathbf{ }\mathbf{}{\mathbf{\text{kg/m}}}^{\mathbf{\text{3}}}$.

1. Length is$L=0.12 \text{m}$
2. Temperature is$T_C=-5^o\text{C}$
3. Thermal conductivity of ice is $k=0.400 \text{W/m.K}$
4. Density of water is$\rho =1000 {\text{kg/m}}^{\text{3}}$

According to Fourier's Law, the amount of time it takes for heat to go through a material is proportional to the negative gradient of the temperature as well as the cross-sectional area that is perpendicular to the gradient. We use the concept of rate of heat conducted. Using the equation of power, we can find the energy conducted. Differentiating the latent heat equation, we get the rate of mass conversion. Using the equation of density, we can write it in terms of area and height. Then, on differentiating it, we can find the rate of length movement.

Formulae:

The rate of conduction due to radiation,

$P_{cond}=\frac{Q}{t}\text{or}\frac{kA\left(T_H-T_C\right)}{L}$ .......(i)

The heat released by the body due to fusion,

$Q=L_{f}m$ .......(ii)

The density of the body,

$\rho =\frac{m}{V}$ .......(iii)

Using equation (i) of conduction rate and the given values, we get the rate of upward conduction as

$\begin{array}{rcl}{P}_{cond}& =& \frac{\left(0.400 \text{W/m.K}\right)A\left(0^{\text{0}}\text{C}-\left(-{5.0}^{\text{0}}\text{C}\right)\right)}{0.12 \text{m}}\\ & =& 16.66A\\ & \approx & \left(16.7A\right)\text{W}\end{array}$

Hence, the conduction rate at upward direction is$\left(16.7A\right)W$ .

Differentiating the equation (ii) with the respect to time, we get the rate of conduction as

$\begin{array}{rcl}{P}_{cond}& =& \frac{d\left(m{L}_f\right)}{dt}\text{(}\because P_{cond}=\frac{dQ}{dt}\text{)}\\ & =& \frac{L_{f}dm}{dt}\end{array}$

We can write the above equation as,

$\frac{dm}{dt}=\frac{P_{cond}}{L_f}$

We know the latent heat of fusion is $L_f=3.33\times {10}^5\text{J/kg}$; plugging the values, we get

$$\begin{gathered} \frac{dm}{dt}=\frac{\left(16.7A\right) \text{W}}{3.33\times {10}^5 \text{J/kg}}·\left(\frac{\text{1J/s}}{1 \text{W}}\right) \\ =\left(5.02\times {10}^{-5}A\right)\text{kg/s} \end{gathered}$$

Hence, the rate of conversion of the mass is$\left(5.02\times {10}^{-5}A\right)\text{kg/s}$

We know $V=Ah$, plugging it in the equation (iii), we get the mass of the water

$m=\rho Ah$ .......(iv)

Here area and density are constant.

Differentiating mass equation (a) with respect to t, we get

$$\begin{gathered} \frac{dm}{dt}=\rho A\left(\frac{dh}{dt}\right) \\ \frac{dh}{dt}=\frac{1}{\rho A}\left(\frac{dm}{dt}\right) \end{gathered}$$

We know the density of water is$\rho =1000 {\text{kg/m}}^{\text{3}}$ ; plugging the given values in the above equation, we get

$\begin{array}{rcl}\frac{dh}{dt}& =& \frac{1}{\left(1000 {\text{kg/m}}^{\text{3}}\right)A}\left(5.02\times {10}^{-5}A\right)\text{kg/s}\\ & =& 5.02\times {10}^{-8}\text{m/s}\end{array}$

Hence, the value of the rate of the height is$5.02\times {10}^{-8}\text{m/s}$