Two liquids at temperature `60^@ C` and `20^@ C` respectively have masses in the ratio `3:4` their specific heat in the ratio `4:5`. If the two liquids are mixed, the resultant temperature is.

Options

A. `70^(@)C`
B. `50^(@)C`
C. `40^(@)C`
D. `35^(@)C`

Explanation

To find the resultant temperature when two liquids are mixed, we can use the formula for the final temperature \( T_f \):

\[ T_f = \frac{m_1 s_1 T_1 + m_2 s_2 T_2}{m_1 s_1 + m_2 s_2} \]

Where:
- \( m_1 \) and \( m_2 \) are the masses of the two liquids.
- \( s_1 \) and \( s_2 \) are the specific heats of the two liquids.
- \( T_1 \) and \( T_2 \) are the initial temperatures of the two liquids.

### Step 1: Assign values based on the given ratios
Let:
- Mass of liquid 1, \( m_1 = 3m \) (where \( m \) is a common mass unit)
- Mass of liquid 2, \( m_2 = 4m \)

From the problem, we have:
- Specific heat of liquid 1, \( s_1 = 4s \)
- Specific heat of liquid 2, \( s_2 = 5s \)

The temperatures are given as:
- \( T_1 = 60^\circ C \)
- \( T_2 = 20^\circ C \)

### Step 2: Substitute the values into the formula
Now, substituting the values into the formula:

\[ T_f = \frac{(3m)(4s)(60) + (4m)(5s)(20)}{(3m)(4s) + (4m)(5s)} \]

### Step 3: Simplify the numerator
Calculating the numerator:

\[ = \frac{(3m)(4s)(60) + (4m)(5s)(20)}{(3m)(4s) + (4m)(5s)} \]

Calculating each term:
- First term: \( 3m \cdot 4s \cdot 60 = 720ms \)
- Second term: \( 4m \cdot 5s \cdot 20 = 400ms \)

So, the numerator becomes:

\[ 720ms + 400ms = 1120ms \]

### Step 4: Simplify the denominator
Calculating the denominator:

\[ = (3m)(4s) + (4m)(5s) = 12ms + 20ms = 32ms \]

### Step 5: Final calculation
Now substituting back into the formula:

\[ T_f = \frac{1120ms}{32ms} \]

The \( ms \) cancels out:

\[ T_f = \frac{1120}{32} = 35^\circ C \]

### Conclusion
The resultant temperature when the two liquids are mixed is \( 35^\circ C \).

---