Question

A cylinder of ice is made inside an insulated cup as shown. The bottom of cup is a plate of thermal conductivity 3.35 W/mK. It's other side is at 100°C. Density of ice = 0.9 gm/cc, density of water = 1 gm/cc. What is the downward speed v of the top surface (in mm/s). L = 3.35 x 10⁵ J/kg.

Answer 1

10/9 mm/s

Answer 2

The rate of heat transfer through the bottom plate is calculated using Fourier's Law of conduction: $\frac{Q}{t} = k A \frac{\Delta T}{\Delta x}$. Given: $k = 3.35$ W/mK $A = \pi r^2 = \pi (0.1 \text{ m})^2 = 0.01\pi \text{ m}^2$ (radius $r = 20\text{cm}/2 = 10\text{cm} = 0.1\text{m}$) $\Delta T = 100^\circ\text{C} - 0^\circ\text{C} = 100$ K $\Delta x = 1 \text{ mm} = 0.001 \text{ m}$

$\frac{Q}{t} = (3.35 \text{ W/mK}) \times (0.01\pi \text{ m}^2) \times \frac{100 \text{ K}}{0.001 \text{ m}} = 3.35 \times 0.01\pi \times 100000 \text{ W} = 3350\pi \text{ W}$

This heat is used to melt the ice. The rate of mass melting ($dm/dt$) is given by $\frac{Q}{t} = \frac{dm}{dt} L$. $\frac{dm}{dt} = \frac{3350\pi \text{ J/s}}{3.35 \times 10^5 \text{ J/kg}} = \frac{3.35 \times 10^3 \pi}{3.35 \times 10^5} \text{ kg/s} = \frac{\pi}{100} \text{ kg/s}$

The rate of volume decrease of ice ($dV_{ice}/dt$) is $\frac{dV_{ice}}{dt} = \frac{dm/dt}{\rho_{ice}}$. Density of ice $\rho_{ice} = 0.9 \text{ gm/cc} = 0.9 \times 10^3 \text{ kg/m}^3$. $\frac{dV_{ice}}{dt} = \frac{\pi/100 \text{ kg/s}}{0.9 \times 10^3 \text{ kg/m}^3} = \frac{\pi}{90000} \text{ m}^3/\text{s}$

The downward speed $v$ of the top surface is related to the rate of volume decrease by $\frac{dV_{ice}}{dt} = A v$. $v = \frac{1}{A} \frac{dV_{ice}}{dt} = \frac{1}{0.01\pi \text{ m}^2} \times \frac{\pi}{90000} \text{ m}^3/\text{s} = \frac{1}{0.01 \times 90000} \text{ m/s} = \frac{1}{900} \text{ m/s}$

Converting to mm/s: $v = \frac{1}{900} \text{ m/s} \times 1000 \text{ mm/m} = \frac{1000}{900} \text{ mm/s} = \frac{10}{9} \text{ mm/s}$