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 water, we will use the principle of conservation of energy, which states that the heat lost by the hot water will be equal to the heat gained by the cold water.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of hot water (m1) = 10 g
- Initial temperature of hot water (T1) = 70 °C
- Mass of cold water (m2) = 5 g
- Initial temperature of cold water (T2) = 30 °C
- Specific heat of water (C) = 1 cal/g°C

2. **Set up the heat transfer equation:**
Since no heat is lost to the surroundings, the heat lost by the hot water is equal to the heat gained by the cold water.
\[ \text{Heat lost by hot water} = \text{Heat gained by cold water} \]
\[ m_1 \cdot C \cdot (T_1 - T) = m_2 \cdot C \cdot (T - T_2) \]

3. **Cancel the specific heat (C) from both sides:**
\[ m_1 \cdot (T_1 - T) = m_2 \cdot (T - T_2) \]

4. **Substitute the values:**
\[ 10 \cdot (70 - T) = 5 \cdot (T - 30) \]

5. **Expand both sides:**
\[ 700 - 10T = 5T - 150 \]

6. **Rearrange the equation to isolate T:**
\[ 700 + 150 = 10T + 5T \]
\[ 850 = 15T \]

7. **Solve for T:**
\[ T = \frac{850}{15} = 56.67 °C \]

### Final Answer:
The equilibrium temperature of the mixture is **56.67 °C**.

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