`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, which states that the heat lost by the coffee will be equal to the heat gained by the cup.

### Step-by-step Solution:

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

2. **Set Up the Heat Transfer Equation:**
The heat lost by the coffee will be equal to the heat gained by the cup:
\[ \text{Heat lost by coffee} = \text{Heat gained by cup} \]
This can be expressed as:
\[ m_2 \cdot s_{coffee} \cdot (T_2 - T) = m_1 \cdot s_{cup} \cdot (T - T_1) \]
where \( T \) is the final equilibrium temperature.

3. **Substitute the Known Values:**
\[ 0.3 \cdot 4080 \cdot (70 - T) = 0.12 \cdot 1020 \cdot (T - 20) \]

4. **Simplify the Equation:**
Calculate the left side:
\[ 0.3 \cdot 4080 = 1224 \quad \Rightarrow \quad 1224 \cdot (70 - T) \]
Calculate the right side:
\[ 0.12 \cdot 1020 = 122.4 \quad \Rightarrow \quad 122.4 \cdot (T - 20) \]
So the equation becomes:
\[ 1224(70 - T) = 122.4(T - 20) \]

5. **Expand Both Sides:**
\[ 1224 \cdot 70 - 1224T = 122.4T - 2448 \]
\[ 85680 - 1224T = 122.4T - 2448 \]

6. **Combine Like Terms:**
Move all terms involving \( T \) to one side and constant terms to the other:
\[ 85680 + 2448 = 1224T + 122.4T \]
\[ 88128 = 1346.4T \]

7. **Solve for \( T \):**
\[ T = \frac{88128}{1346.4} \approx 65.5^\circ C \]

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