A beaker contains 200 g of water. The heat capacity of the beaker is equal to that of 20 g of water. The initial temperature of water in the beaker is `20^@C` .If 440 g of hot water at `92^@C` is poured in it, the final temperature (neglecting radiation loss) will be nearest to

Options

A. `58^(@)C`
B. `68^(@)C`
C. `73^(@)C`
D. `78^(@)C`

Explanation

To solve the problem, 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 cooler water and the beaker.

### Step-by-step Solution:

1. **Identify the Variables:**
- Mass of cold water (m1) = 200 g
- Mass of hot water (m2) = 440 g
- Mass equivalent of the beaker (m3) = 20 g (since its heat capacity is equal to that of 20 g of water)
- Initial temperature of cold water (T1) = 20 °C
- Initial temperature of hot water (T2) = 92 °C
- Final temperature (Tf) = ?

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 and the beaker:
\[ \text{Heat lost by hot water} = \text{Heat gained by cold water} + \text{Heat gained by beaker} \]
Mathematically, this can be expressed as:
\[ m2 \cdot c \cdot (T2 - Tf) = m1 \cdot c \cdot (Tf - T1) + m3 \cdot c \cdot (Tf - T1) \]
Here, \(c\) is the specific heat capacity of water, which we can cancel out from both sides since it is the same for all terms.

3. **Substitute the Known Values:**
\[ 440 \cdot (92 - Tf) = 200 \cdot (Tf - 20) + 20 \cdot (Tf - 20) \]
Simplifying the right side:
\[ 440 \cdot (92 - Tf) = 200 \cdot (Tf - 20) + 20 \cdot (Tf - 20) \]
\[ = 200Tf - 4000 + 20Tf - 400 \]
\[ = 220Tf - 4400 \]

4. **Expand and Rearrange the Equation:**
\[ 440 \cdot 92 - 440Tf = 220Tf - 4400 \]
\[ 40480 - 440Tf = 220Tf - 4400 \]
Combine like terms:
\[ 40480 + 4400 = 440Tf + 220Tf \]
\[ 44880 = 660Tf \]

5. **Solve for Tf:**
\[ Tf = \frac{44880}{660} \approx 68 °C \]

### Final Answer:
The final temperature \(Tf\) is approximately **68 °C**.