Calculate the mass of alcohol flowing in two minutes through a tube of radius `5 xx 10^(-4)` m and length 0.5 m, if there is a constant pressure head of 0.6 m of alcohol. Density of alcohol is `800 kg//m^(3)` Coefficient of viscosity of alcohol =` 1.38 xx 10^(-3) Ns//m^(3)` ?

Explanation

To solve the problem of calculating the mass of alcohol flowing through a tube under the given conditions, we can follow these steps:

### Step 1: Calculate the Pressure
The pressure due to the height of the alcohol column can be calculated using the formula:
\[ P = h \cdot \rho \cdot g \]
where:
- \( h = 0.6 \, \text{m} \) (height of the alcohol column)
- \( \rho = 800 \, \text{kg/m}^3 \) (density of alcohol)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ P = 0.6 \cdot 800 \cdot 9.8 \]
\[ P = 4704 \, \text{Pa} \]

### Step 2: Calculate the Volume Flow Rate (Q)
Using Poiseuille's Law, the volume flow rate \( Q \) can be calculated as:
\[ Q = \frac{\pi r^4 P}{8 \eta L} \]
where:
- \( r = 5 \times 10^{-4} \, \text{m} \) (radius of the tube)
- \( \eta = 1.38 \times 10^{-3} \, \text{Ns/m}^2 \) (viscosity of alcohol)
- \( L = 0.5 \, \text{m} \) (length of the tube)

Substituting the values:
\[ Q = \frac{\pi (5 \times 10^{-4})^4 \cdot 4704}{8 \cdot (1.38 \times 10^{-3}) \cdot 0.5} \]

Calculating \( (5 \times 10^{-4})^4 \):
\[ (5 \times 10^{-4})^4 = 6.25 \times 10^{-16} \]

Now substituting this back into the equation:
\[ Q = \frac{\pi \cdot 6.25 \times 10^{-16} \cdot 4704}{8 \cdot 1.38 \times 10^{-3} \cdot 0.5} \]
\[ Q \approx \frac{9.23 \times 10^{-12}}{5.52 \times 10^{-4}} \]
\[ Q \approx 1.67 \times 10^{-7} \, \text{m}^3/\text{s} \]

### Step 3: Calculate the Total Volume in 2 Minutes
Convert 2 minutes to seconds:
\[ t = 2 \times 60 = 120 \, \text{s} \]

Now calculate the total volume \( V \):
\[ V = Q \cdot t \]
\[ V = 1.67 \times 10^{-7} \cdot 120 \]
\[ V \approx 2.00 \times 10^{-5} \, \text{m}^3 \]

### Step 4: Calculate the Mass of Alcohol
Using the formula for mass:
\[ m = V \cdot \rho \]
Substituting the values:
\[ m = 2.00 \times 10^{-5} \cdot 800 \]
\[ m \approx 1.6 \times 10^{-2} \, \text{kg} \]

### Final Answer
The mass of alcohol flowing through the tube in 2 minutes is approximately:
\[ m \approx 0.016 \, \text{kg} \text{ or } 16 \, \text{grams} \]