Three liquids with masses `m_1`,`m_2`,`m_3` are throughly mixed. If their specific heats are `c_1`,`c_2`,`c_3` and their temperature `T_1`,`T_2`,`T_3`, respectively, then the temperature of the mixture is

Options

A. `(c_1T_1+c_2T_2+c_3T_3)/(m_1c_1+m_2c_2+m_3c_3)`
B. `(m_1c_1 T_1+m_2c_2t_2+m_3c_3T_3)/(m_1c_1+m_2c_2+m_3c_3)`
C. `(m_1c_1T_1+m_2c_2T_2+m_3c_3T_3)/(m_1T_1+m_2T_2+m_3T_3)`
D. `(m_1T_1+m_2T_2+m_3T_3)/(c_1T_1+c_2T_2+c_3T_3)`

Explanation

To find the final temperature of a mixture of three liquids with given masses, specific heats, and initial temperatures, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let the masses of the three liquids be \( m_1, m_2, m_3 \).
- Let the specific heats of the three liquids be \( c_1, c_2, c_3 \).
- Let the initial temperatures of the three liquids be \( T_1, T_2, T_3 \).

2. **Calculate the Total Heat Supplied**:
- When the liquids are cooled down to 0°C, the heat lost by each liquid can be expressed as:
- For liquid 1: \( Q_1 = m_1 c_1 (T_1 - 0) = m_1 c_1 T_1 \)
- For liquid 2: \( Q_2 = m_2 c_2 (T_2 - 0) = m_2 c_2 T_2 \)
- For liquid 3: \( Q_3 = m_3 c_3 (T_3 - 0) = m_3 c_3 T_3 \)
- Therefore, the total heat supplied by all three liquids when they cool down to 0°C is:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 = m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3 \]

3. **Calculate the Heat Used to Raise the Temperature**:
- After mixing, the temperature of the mixture rises from 0°C to a final temperature \( T \).
- The heat used to raise the temperature of each liquid from 0°C to \( T \) can be expressed as:
- For liquid 1: \( Q'_1 = m_1 c_1 (T - 0) = m_1 c_1 T \)
- For liquid 2: \( Q'_2 = m_2 c_2 (T - 0) = m_2 c_2 T \)
- For liquid 3: \( Q'_3 = m_3 c_3 (T - 0) = m_3 c_3 T \)
- Therefore, the total heat used to raise the temperature to \( T \) is:
\[ Q'_{\text{total}} = Q'_1 + Q'_2 + Q'_3 = m_1 c_1 T + m_2 c_2 T + m_3 c_3 T \]

4. **Apply the Principle of Conservation of Energy**:
- According to the law of calorimetry, the total heat lost by the liquids while cooling down to 0°C equals the total heat gained while raising the temperature to \( T \):
\[ m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3 = m_1 c_1 T + m_2 c_2 T + m_3 c_3 T \]

5. **Rearrange the Equation**:
- Rearranging the equation gives:
\[ m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3 = T (m_1 c_1 + m_2 c_2 + m_3 c_3) \]

6. **Solve for the Final Temperature \( T \)**:
- To find \( T \), divide both sides by the total heat capacity:
\[ T = \frac{m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3}{m_1 c_1 + m_2 c_2 + m_3 c_3} \]

### Final Answer:
The final temperature \( T \) of the mixture is given by:
\[ T = \frac{m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3}{m_1 c_1 + m_2 c_2 + m_3 c_3} \]