Two liquids are at temperatures `20^@C and 40^@C`. When same mass of both of them is mixed, the temperature of the mixture is `32^@C`. What is the ratio of their specific heats?

Options

A. `1//3`
B. `2//5`
C. `3//2`
D. `2//3`

Explanation

To solve the problem, we need to find the ratio of the specific heats of two liquids mixed together. Let's denote the specific heat of liquid 1 as \( S_1 \) and that of liquid 2 as \( S_2 \). The temperatures of the two liquids before mixing are given as follows:

- Temperature of liquid 1, \( T_1 = 20^\circ C \)
- Temperature of liquid 2, \( T_2 = 40^\circ C \)
- Final temperature of the mixture, \( T_f = 32^\circ C \)

### Step-by-Step Solution:

1. **Identify the heat gained and lost:**
- Liquid 1 (at \( 20^\circ C \)) gains heat as it warms up to \( 32^\circ C \).
- Liquid 2 (at \( 40^\circ C \)) loses heat as it cools down to \( 32^\circ C \).

2. **Calculate the change in temperature for both liquids:**
- For liquid 1:
\[ \Delta T_1 = T_f - T_1 = 32^\circ C - 20^\circ C = 12^\circ C \]
- For liquid 2:
\[ \Delta T_2 = T_2 - T_f = 40^\circ C - 32^\circ C = 8^\circ C \]

3. **Write the heat gained and lost equations:**
- Heat gained by liquid 1:
\[ Q_1 = m \cdot S_1 \cdot \Delta T_1 = m \cdot S_1 \cdot 12 \]
- Heat lost by liquid 2:
\[ Q_2 = m \cdot S_2 \cdot \Delta T_2 = m \cdot S_2 \cdot 8 \]

4. **Set the heat gained equal to the heat lost:**
Since the heat gained by liquid 1 is equal to the heat lost by liquid 2, we have:
\[ Q_1 = Q_2 \]
Therefore:
\[ m \cdot S_1 \cdot 12 = m \cdot S_2 \cdot 8 \]

5. **Cancel out the mass \( m \):**
Since the masses are the same, we can cancel \( m \) from both sides:
\[ S_1 \cdot 12 = S_2 \cdot 8 \]

6. **Rearrange the equation to find the ratio of specific heats:**
\[ \frac{S_1}{S_2} = \frac{8}{12} = \frac{2}{3} \]

### Final Answer:
The ratio of the specific heats of the two liquids is:
\[ \frac{S_1}{S_2} = \frac{2}{3} \]