A liquid of mass m and specific heat c is heated to a temperature 2T. Another liquid of mass m/2 and specific heat 2 c is heated to a temperature T. If these two liquids are mixed, the resulting temperature of the mixture is

Options

A. (2/3) T
B. (8/5)T
C. (3/5)T
D. `(3/2)`T

Explanation

To find the resulting temperature of the mixture when two liquids are combined, we can use the principle of conservation of energy. The heat lost by the hotter liquid will be equal to the heat gained by the cooler liquid.

### Step-by-Step Solution:

1. **Identify the parameters of the two liquids:**
- For the first liquid:
- Mass, \( m_1 = m \)
- Specific heat, \( c_1 = c \)
- Initial temperature, \( T_1 = 2T \)
- For the second liquid:
- Mass, \( m_2 = \frac{m}{2} \)
- Specific heat, \( c_2 = 2c \)
- Initial temperature, \( T_2 = T \)

2. **Set up the equation for the final temperature \( T_f \):**
The final temperature \( T_f \) can be calculated using the formula:
\[ T_f = \frac{m_1 c_1 T_1 + m_2 c_2 T_2}{m_1 c_1 + m_2 c_2} \]

3. **Substitute the values into the equation:**
- Substitute \( m_1, c_1, T_1, m_2, c_2, T_2 \):
\[ T_f = \frac{m \cdot c \cdot (2T) + \frac{m}{2} \cdot (2c) \cdot T}{m \cdot c + \frac{m}{2} \cdot (2c)} \]

4. **Simplify the numerator:**
- Calculate the numerator:
\[ = \frac{2mcT + mcT}{m c + mc} = \frac{3mcT}{mc + mc} = \frac{3mcT}{2mc} \]

5. **Simplify the denominator:**
- The denominator simplifies to:
\[ = 2mc \]

6. **Final calculation:**
- Now substitute back into the equation:
\[ T_f = \frac{3mcT}{2mc} = \frac{3}{2}T \]

7. **Conclusion:**
- The resulting temperature of the mixture is:
\[ T_f = \frac{3}{2} T \]

### Final Answer:
The resulting temperature of the mixture is \( \frac{3}{2} T \).