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, 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 masses and temperatures:**
- Mass of hot water, \( m_1 = 100 \, \text{g} \)
- Temperature of hot water, \( T_1 = 90^\circ C \)
- Mass of cold water, \( m_2 = 600 \, \text{g} \)
- Temperature of cold water, \( T_2 = 20^\circ C \)

2. **Set up the heat balance equation:**
The heat lost by the hot water will be 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. **Substituting the values:**
\[ 100 (90 - T) = 600 (T - 20) \]

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

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

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

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

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