Question

How much water should be evaporated from 1 kg milk of 9% concentration in order to form a 36% solution?

• A. 0.325 kg
• B. 0.250 kg
• C. 0.750 kg
• D. 0.625 kg

Hint:

Always calculate the final mass first using solids conservation: $m_2 = m_1 \cdot (x_1 / x_2)$. Then, subtract the final mass from the initial mass to find the evaporated water: $1 - (9/36) = 1 - 0.25 = 0.750$ kg.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Concentration processes in dairy engineering, such as evaporation, involve removing water from milk while retaining the milk solids.
To calculate the amount of water to be evaporated, a mass balance must be performed on the total product mass and the total solids content.
The total mass of milk solids remains constant during the concentration process.
Key Formula or Approach:
The mass balance equation for milk solids is:
\[ m_1 \cdot x_1 = m_2 \cdot x_2 \]
Where:
$m_1$ is the initial mass of the milk.
$x_1$ is the initial concentration of solids (expressed as a fraction).
$m_2$ is the final mass of the concentrated solution.
$x_2$ is the final concentration of solids (expressed as a fraction).
The mass of water evaporated ($m_w$) is:
\[ m_w = m_1 - m_2 \]

Step 2: Detailed Explanation:
Let us substitute the given values into the mass balance equations:
The initial mass of milk is:
\[ m_1 = 1 \text{ kg} \]
The initial solids concentration is 9%, which can be written as:
\[ x_1 = 0.09 \]
The target concentrated solids concentration is 36%, which can be written as:
\[ x_2 = 0.36 \]
Using the solids mass balance:
\[ 1 \cdot 0.09 = m_2 \cdot 0.36 \]
Solving for the final concentrated mass ($m_2$):
\[ m_2 = \frac{0.09}{0.36} \]
\[ m_2 = 0.250 \text{ kg} \]
Now, we calculate the mass of water that must be evaporated ($m_w$):
\[ m_w = m_1 - m_2 \]
\[ m_w = 1 - 0.250 \]
\[ m_w = 0.750 \text{ kg} \]
Therefore, 0.750 kg of water must be evaporated to concentrate the milk solids.

Step 3: Final Answer
The amount of water to be evaporated is 0.750 kg.