Milk has a viscosity of about 2.1 cp at 20 C. The density of whole milk is about 1.03 g/cc and milk fat (cream) has a density of about .93 g/cc and the average drop of cream has an 8 μ diameter. Use Stokes's law to calculate how long it will take a drop of cream to rise 10 cm. O 12 hr 40 min O 16 hr 45 min 6000 sec O 10 hr 45 min
The Correct Answer and Explanation is
The correct answer is 16 hours 45 minutes.
Explanation:
To determine how long it takes for a drop of cream (fat) to rise through milk, we use Stokes’ Law, which applies to small spheres moving through a viscous fluid. The formula for the terminal velocity of a small sphere in a fluid is:
[
v = \frac{2}{9} \frac{r^2 g (\rho_m – \rho_c)}{\mu}
]
where:
- ( v ) = terminal velocity (m/s)
- ( r ) = radius of the drop (m)
- ( g ) = gravitational acceleration (( 9.81 ) m/s²)
- ( \rho_m ) = density of milk (( 1.03 ) g/cc = ( 1030 ) kg/m³)
- ( \rho_c ) = density of cream (( 0.93 ) g/cc = ( 930 ) kg/m³)
- ( \mu ) = viscosity of milk (( 2.1 ) cP = ( 2.1 \times 10^{-3} ) Pa·s)
The given drop of cream has a diameter of 8 μm (or ( 8 \times 10^{-6} ) m), so its radius is:
[
r = \frac{8 \times 10^{-6}}{2} = 4 \times 10^{-6} \text{ m}
]
Plugging in the values:
[
v = \frac{2}{9} \times \frac{(4 \times 10^{-6})^2 \times 9.81 \times (1030 – 930)}{2.1 \times 10^{-3}}
]
[
v = 1.66 \times 10^{-6} \text{ m/s}
]
To rise 10 cm (0.1 m):
[
t = \frac{0.1}{1.66 \times 10^{-6}} = 60226.99 \text{ seconds}
]
Converting:
[
60226.99 \text{ s} = 16 \text{ hours } 45 \text{ minutes}
]
Thus, the correct answer is 16 hours 45 minutes.