A steel bar in the form of a rectangular parallelepiped of height $a$, breadth $b$, and length $c$ is embedded in a cake of ice as shown in Fig. PII-2. With the aid of an external magnetic field, a constant force $F$ is exerted downward on the bar. The whole system is at $0^{\circ} \mathrm{C}$.
(a) Show that the decrease in temperature of the ice directly below the bar is
$$
\Delta T=\frac{F T\left(v^{t}-v^{\prime \prime}\right)}{b c \Delta h_{F}}
$$
(b) Ice melts (see Prob. $11.12$ ) under the bar, and all the water thus formed is forced to the top of the bar, where it refreezes. This phenomenon is known as regelation. Heat, therefore, is liberated above the bar, is conducted through the metal and a layer of water under the metal, and is absorbed by the ice under
the layer of water. Show that the speed with which the bar sinks through the ice is
$$
\frac{d y}{d t}=\frac{U^{\prime} T\left(v^{\prime}-v^{\prime \prime}\right) F}{\rho\left(\Delta h_{F}\right)^{2} b c}
$$
where $U^{\prime}$ is the overall heat-transfer coefficient of the composite heat-conducting path consisting of the metal and the water layer. $U^{\prime}$ is given by
$$
\frac{1}{U^{\prime}}=\frac{x_{m}}{K_{m}}+\frac{x_{w}}{K_{w}}
$$
where $x_{m}$ and $x_{w}$ are the thicknesses of the metal and water layer, respectively, and $K_{m}$ and $K_{w}$ are their respective thermal conductivities.
(c) Assuming that the water layer has a thickness of about $10^{-5} \mathrm{~m}$ and a thermal conductivity of about $0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, and that the bar is $0.1 \mathrm{~m}$ long, with $a$ and $b$ each equal to $10^{-3} \mathrm{~m}$, with what speed will the bar descend when $F=10^{2} \mathrm{~N} ?$ (Thermal conductivity of stecl is $60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .$ )