Two liquid A and B are at `32^@C` and `24^@C`. When mixed in equal masses the temperature of the mixture is found to be `28^@C`. Their specific heats are in the ratio of

Options

A. `3:2`
B. `2:3`
C. `1:1`
D. `4:3`

Explanation

To solve the problem of finding the ratio of specific heats of two liquids A and B, we can follow these steps:

### Step 1: Understand the problem
We have two liquids A and B at different initial temperatures (32°C and 24°C respectively). When mixed in equal masses, the final temperature of the mixture is 28°C. We need to find the ratio of their specific heats.

### Step 2: Set up the heat transfer equation
According to the principle of calorimetry, the heat lost by the hotter liquid (A) will be equal to the heat gained by the cooler liquid (B). We can express this mathematically as:
\[ \text{Heat lost by A} = \text{Heat gained by B} \]

### Step 3: Write the expressions for heat lost and gained
The heat lost by liquid A can be expressed as:
\[ Q_A = m \cdot S_A \cdot (T_{initial, A} - T_{final}) \]
Where:
- \( m \) = mass of liquid A
- \( S_A \) = specific heat of liquid A
- \( T_{initial, A} = 32°C \)
- \( T_{final} = 28°C \)

Thus,
\[ Q_A = m \cdot S_A \cdot (32 - 28) = m \cdot S_A \cdot 4 \]

The heat gained by liquid B can be expressed as:
\[ Q_B = m \cdot S_B \cdot (T_{final} - T_{initial, B}) \]
Where:
- \( S_B \) = specific heat of liquid B
- \( T_{initial, B} = 24°C \)

Thus,
\[ Q_B = m \cdot S_B \cdot (28 - 24) = m \cdot S_B \cdot 4 \]

### Step 4: Set the heat lost equal to the heat gained
From the principle of conservation of energy, we can equate the two expressions:
\[ m \cdot S_A \cdot 4 = m \cdot S_B \cdot 4 \]

### Step 5: Simplify the equation
Since the masses are equal and the factor of 4 is common on both sides, we can cancel them out:
\[ S_A = S_B \]

### Step 6: Find the ratio of specific heats
Thus, the ratio of specific heats \( S_A \) to \( S_B \) is:
\[ \frac{S_A}{S_B} = 1 \]

### Conclusion
The specific heats of liquids A and B are in the ratio of 1:1.

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