10 g of water at `70^@C` is mixed with 5 g of water at `30^@C`. Find the temperature of the mixture in equilibrium. Specific heat of water is `1 cal//g-^@C`.

Explanation

To find the equilibrium temperature of the mixture of two water samples, we can use the principle of conservation of energy. The heat lost by the hotter water will be equal to the heat gained by the cooler water.

### Step-by-Step Solution:

1. **Identify the known values**:
- Mass of water at 70°C, \( m_1 = 10 \, \text{g} \)
- Initial temperature of water at 70°C, \( T_1 = 70 \, \text{°C} \)
- Mass of water at 30°C, \( m_2 = 5 \, \text{g} \)
- Initial temperature of water at 30°C, \( T_2 = 30 \, \text{°C} \)
- Specific heat of water, \( c = 1 \, \text{cal/g°C} \)

2. **Set up the heat transfer equation**:
The heat lost by the hot water will be equal to the heat gained by the cold water:
\[ m_1 \cdot c \cdot (T_1 - T) = m_2 \cdot c \cdot (T - T_2) \]
Here, \( T \) is the final equilibrium temperature.

3. **Substitute the known values into the equation**:
\[ 10 \cdot 1 \cdot (70 - T) = 5 \cdot 1 \cdot (T - 30) \]

4. **Simplify the equation**:
\[ 10(70 - T) = 5(T - 30) \]
Expanding both sides gives:
\[ 700 - 10T = 5T - 150 \]

5. **Rearrange the equation to solve for \( T \)**:
Combine like terms:
\[ 700 + 150 = 10T + 5T \]
\[ 850 = 15T \]

6. **Solve for \( T \)**:
\[ T = \frac{850}{15} = 56.67 \, \text{°C} \]

### Final Answer:
The temperature of the mixture in equilibrium is approximately \( 56.67 \, \text{°C} \).