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 calorimetry, which states that the heat lost by the hot water will be equal to the heat gained by the cold water and the beaker.

### Step-by-Step Solution:

1. **Identify the masses and initial temperatures:**
- Mass of cold water (m1) = 200 g
- Mass of hot water (m2) = 440 g
- Water equivalent of the beaker (m3) = 20 g
- Initial temperature of cold water (T1) = 20°C
- Initial temperature of hot water (T2) = 92°C

2. **Define the final temperature:**
- Let the final temperature be T (in °C).

3. **Write the heat lost and gained equations:**
- Heat lost by hot water:
\[ Q_{\text{lost}} = m2 \cdot c \cdot (T2 - T) = 440 \cdot 1 \cdot (92 - T) \]
- Heat gained by cold water:
\[ Q_{\text{gained}} = m1 \cdot c \cdot (T - T1) = 200 \cdot 1 \cdot (T - 20) \]
- Heat gained by the beaker:
\[ Q_{\text{gained, beaker}} = m3 \cdot c \cdot (T - T1) = 20 \cdot 1 \cdot (T - 20) \]

4. **Set up the equation based on the principle of calorimetry:**
\[ Q_{\text{lost}} = Q_{\text{gained}} + Q_{\text{gained, beaker}} \]
\[ 440 \cdot (92 - T) = 200 \cdot (T - 20) + 20 \cdot (T - 20) \]

5. **Combine the heat gained terms:**
\[ 440 \cdot (92 - T) = (200 + 20) \cdot (T - 20) \]
\[ 440 \cdot (92 - T) = 220 \cdot (T - 20) \]

6. **Expand both sides:**
\[ 440 \cdot 92 - 440T = 220T - 4400 \]

7. **Rearrange the equation:**
\[ 440 \cdot 92 + 4400 = 220T + 440T \]
\[ 440 \cdot 92 + 4400 = 660T \]

8. **Calculate the left side:**
\[ 440 \cdot 92 = 40480 \]
\[ 40480 + 4400 = 44880 \]
\[ 44880 = 660T \]

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

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