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

Answer

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

### Step 1: Define the Variables
Let:
- \( x \) = mass of water taken from tank A (at \( 30^\circ C \))
- \( 40 - x \) = mass of water taken from tank B (at \( 80^\circ C \))
- Total mass of water = \( 40 \) kg

### Step 2: Set Up the Heat Transfer Equation
According to the principle of calorimetry, the heat lost by the hotter water (from tank B) will equal the heat gained by the cooler water (from tank A).

The heat lost by water from tank B can be expressed as:
\[ Q_{\text{loss}} = (40 - x) \cdot c \cdot (T_B - T_f) \]
Where:
- \( c \) = specific heat capacity of water (which is constant and can be taken as 1 for simplicity)
- \( T_B = 80^\circ C \) (initial temperature of tank B)
- \( T_f = 50^\circ C \) (final temperature)

So,
\[ Q_{\text{loss}} = (40 - x) \cdot 1 \cdot (80 - 50) = (40 - x) \cdot 30 \]

The heat gained by water from tank A can be expressed as:
\[ Q_{\text{gain}} = x \cdot c \cdot (T_f - T_A) \]
Where:
- \( T_A = 30^\circ C \) (initial temperature of tank A)

So,
\[ Q_{\text{gain}} = x \cdot 1 \cdot (50 - 30) = x \cdot 20 \]

### Step 3: Set the Heat Loss Equal to Heat Gain
Now, we can set the two equations equal to each other:
\[ (40 - x) \cdot 30 = x \cdot 20 \]

### Step 4: Solve for \( x \)
Expanding the equation:
\[ 1200 - 30x = 20x \]
Combining like terms:
\[ 1200 = 50x \]
Dividing both sides by 50:
\[ x = \frac{1200}{50} = 24 \text{ kg} \]

### Step 5: Calculate the Mass from Tank B
Now, we can find the mass of water taken from tank B:
\[ 40 - x = 40 - 24 = 16 \text{ kg} \]

### Final Answer
The amounts of water taken from each tank are:
- From tank A: \( 24 \) kg
- From tank B: \( 16 \) kg