`0.3 Kg` of hot coffee, which is at `70^(@)C`, is poured into a cup of mass 0.12 kg . Find the final equilibrium temperature. Take room temperature at `20^(@)C` .
(`s_(coffee) = 4080 J/kg-K and s_(cup) = 1020 J/kg-K`.)

Options

A. `45.5^(@)C`
B. `55.5^(@)C`
C. `65.5^(@)C`
D. `40.5^(@)C`

Explanation

To find the final equilibrium temperature when hot coffee is poured into a cup, we can use the principle of conservation of energy. The heat lost by the coffee will be equal to the heat gained by the cup.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of coffee, \( m_1 = 0.3 \, \text{kg} \)
- Initial temperature of coffee, \( T_1 = 70 \, ^\circ C \)
- Specific heat of coffee, \( s_1 = 4080 \, \text{J/kg-K} \)
- Mass of cup, \( m_2 = 0.12 \, \text{kg} \)
- Initial temperature of cup, \( T_2 = 20 \, ^\circ C \)
- Specific heat of cup, \( s_2 = 1020 \, \text{J/kg-K} \)

2. **Set up the heat transfer equation:**
The heat lost by the coffee will be equal to the heat gained by the cup:
\[ Q_{\text{coffee}} = Q_{\text{cup}} \]
This can be expressed as:
\[ m_1 s_1 (T_1 - T_f) = m_2 s_2 (T_f - T_2) \]
where \( T_f \) is the final equilibrium temperature.

3. **Substitute the known values into the equation:**
\[ 0.3 \times 4080 \times (70 - T_f) = 0.12 \times 1020 \times (T_f - 20) \]

4. **Simplify the equation:**
Calculate the left side:
\[ 1224 \times (70 - T_f) = 0.12 \times 1020 \times (T_f - 20) \]
The right side simplifies to:
\[ 1224 \times (70 - T_f) = 122.4 \times (T_f - 20) \]

5. **Distribute and rearrange the equation:**
Expanding both sides gives:
\[ 85740 - 1224 T_f = 122.4 T_f - 2448 \]
Rearranging terms leads to:
\[ 85740 + 2448 = 1224 T_f + 122.4 T_f \]
\[ 88188 = 1346.4 T_f \]

6. **Solve for \( T_f \):**
\[ T_f = \frac{88188}{1346.4} \approx 65.5 \, ^\circ C \]

### Final Answer:
The final equilibrium temperature \( T_f \) is approximately \( 65.5 \, ^\circ C \).