Two tanks A and B contain water at `30^@C` and `80^@C` respectively calculate the amount of water that must be taken from each tank respectively to prepare 40 kg of water at `50^@C`

Options

A. 24 kg, 16 kg
B. kg, 24 kg
C. 20 kg, 20 kg
D. 30 kg, 10 kg

Explanation

To solve the problem of mixing water from two tanks at different temperatures to achieve a desired final temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( x \) be the mass of water taken from tank A (at \( 30^\circ C \)).
- Let \( y \) be the mass of water taken from tank B (at \( 80^\circ C \)).
- We know the total mass of water required is \( 40 \, \text{kg} \):
\[ x + y = 40 \quad \text{(Equation 1)} \]

2. **Set Up the Heat Transfer Equation**:
- The heat gained by the water from tank A must equal the heat lost by the water from tank B.
- The heat gained by tank A is given by:
\[ \text{Heat gained} = x \cdot c \cdot (T_f - T_A) = x \cdot c \cdot (50 - 30) \]
- The heat lost by tank B is given by:
\[ \text{Heat lost} = y \cdot c \cdot (T_B - T_f) = y \cdot c \cdot (80 - 50) \]
- Since the specific heat \( c \) is the same for both tanks, it cancels out from the equation:
\[ x \cdot (50 - 30) = y \cdot (80 - 50) \]
- Simplifying this gives:
\[ x \cdot 20 = y \cdot 30 \quad \text{(Equation 2)} \]

3. **Express \( x \) in terms of \( y \)**:
- From Equation 2:
\[ x = \frac{30}{20} y = \frac{3}{2} y \]

4. **Substitute \( x \) in Equation 1**:
- Substitute \( x \) in Equation 1:
\[ \frac{3}{2} y + y = 40 \]
- Combine terms:
\[ \frac{5}{2} y = 40 \]

5. **Solve for \( y \)**:
- Multiply both sides by \( \frac{2}{5} \):
\[ y = 40 \cdot \frac{2}{5} = 16 \, \text{kg} \]

6. **Find \( x \)**:
- Substitute \( y \) back into the expression for \( x \):
\[ x = \frac{3}{2} \cdot 16 = 24 \, \text{kg} \]

### Final Answer:
- The amount of water taken from tank A is \( 24 \, \text{kg} \).
- The amount of water taken from tank B is \( 16 \, \text{kg} \).