Equal masses of two liquid- one at `20^(@)C` and other at `40^(@)C` are mixed together.The temperature of the mixture is `32^(@)C`. The ratio of their specific heats is

Options

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

Explanation

To solve the problem of finding the ratio of specific heats of two liquids mixed together, we can use the principle of calorimetry, which states that the heat gained by the cooler liquid is equal to the heat lost by the warmer liquid.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let the mass of each liquid be \( m \).
- Let the specific heat of the liquid at \( 20^\circ C \) be \( c_1 \).
- Let the specific heat of the liquid at \( 40^\circ C \) be \( c_2 \).
- The initial temperature of the first liquid \( T_1 = 20^\circ C \).
- The initial temperature of the second liquid \( T_2 = 40^\circ C \).
- The final temperature of the mixture \( T = 32^\circ C \).

2. **Apply the Principle of Calorimetry:**
According to the principle of calorimetry:
\[ \text{Heat gained by the cooler liquid} = \text{Heat lost by the warmer liquid} \]
This can be expressed mathematically as:
\[ m c_1 (T - T_1) = m c_2 (T_2 - T) \]
Here, \( T - T_1 \) is the change in temperature for the cooler liquid, and \( T_2 - T \) is the change in temperature for the warmer liquid.

3. **Substitute the Values:**
Substitute the known values into the equation:
\[ m c_1 (32 - 20) = m c_2 (40 - 32) \]
Simplifying this gives:
\[ m c_1 (12) = m c_2 (8) \]

4. **Cancel Mass \( m \):**
Since the masses are equal and non-zero, we can cancel \( m \) from both sides:
\[ c_1 \cdot 12 = c_2 \cdot 8 \]

5. **Rearrange to Find the Ratio:**
Rearranging the equation gives:
\[ \frac{c_1}{c_2} = \frac{8}{12} \]
Simplifying this ratio:
\[ \frac{c_1}{c_2} = \frac{2}{3} \]

6. **Conclusion:**
The ratio of the specific heats of the two liquids is:
\[ c_1 : c_2 = 2 : 3 \]

### Final Answer:
The ratio of their specific heats is \( 2 : 3 \).