Initially , a bearker has 100 g of water at temperature `90^@C` Later another 600 g of water at temperatures `20^@C` was poured into the beaker. The temperature ,T of the water after mixing is

Options

A. `20^@C`
B. `30^@C`
C. `45^@C`
D. `55^@C`

Answer

To find the final temperature \( T \) of the water after mixing, we can use the principle of conservation of energy. 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 values:**
- Mass of hot water, \( m_1 = 100 \, \text{g} \)
- Initial temperature of hot water, \( T_1 = 90^\circ C \)
- Mass of cold water, \( m_2 = 600 \, \text{g} \)
- Initial temperature of cold water, \( T_2 = 20^\circ C \)

2. **Set up the heat transfer equation:**
The heat lost by the hot water is equal to the heat gained by the cold water:
\[ m_1 c (T_1 - T) = m_2 c (T - T_2) \]
Here, \( c \) is the specific heat capacity of water, which cancels out from both sides.

3. **Substitute the values:**
\[ 100 (90 - T) = 600 (T - 20) \]

4. **Expand both sides:**
\[ 9000 - 100T = 600T - 12000 \]

5. **Rearrange the equation:**
Combine like terms:
\[ 9000 + 12000 = 600T + 100T \]
\[ 21000 = 700T \]

6. **Solve for \( T \):**
\[ T = \frac{21000}{700} = 30^\circ C \]

### Final Answer:
The final temperature \( T \) of the water after mixing is \( 30^\circ C \).