A 0.20 kg lead ball is heated to `90.0^(@)C` and dropped into an ideal calorimeter containing 0.50 kg of water initially at `20.0^(@)C`. What is the final equilibrium temperature of the lead ball ? The specific heat capacity of lead is `128 J //( kg. ""^(@)C )`, and the specific heat of water is `4186 J // ( kg. ""^(@)C )`.

Options

A. `4.8^(@)C `
B. `20.8^(@)C`
C. `22.4^(@)C`
D. `27.8^(@)C`

Answer

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the lead ball will be equal to the heat gained by the water.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of lead ball, \( m_{lead} = 0.20 \, \text{kg} \)
- Initial temperature of lead ball, \( T_{initial, lead} = 90.0 \, ^\circ C \)
- Mass of water, \( m_{water} = 0.50 \, \text{kg} \)
- Initial temperature of water, \( T_{initial, water} = 20.0 \, ^\circ C \)
- Specific heat capacity of lead, \( c_{lead} = 128 \, \text{J/(kg} \cdot ^\circ C) \)
- Specific heat capacity of water, \( c_{water} = 4186 \, \text{J/(kg} \cdot ^\circ C) \)

2. **Set Up the Heat Transfer Equation:**
The heat lost by the lead ball will be equal to the heat gained by the water:
\[ Q_{lost} = Q_{gained} \]
This can be expressed as:
\[ m_{lead} \cdot c_{lead} \cdot (T_{initial, lead} - T_f) = m_{water} \cdot c_{water} \cdot (T_f - T_{initial, water}) \]
where \( T_f \) is the final equilibrium temperature.

3. **Substitute the Known Values:**
\[ 0.20 \cdot 128 \cdot (90.0 - T_f) = 0.50 \cdot 4186 \cdot (T_f - 20.0) \]

4. **Simplify the Equation:**
Calculate the left side:
\[ 25.6 \cdot (90.0 - T_f) = 0.50 \cdot 4186 \cdot (T_f - 20.0) \]
Calculate the right side:
\[ 25.6 \cdot (90.0 - T_f) = 2093 \cdot (T_f - 20.0) \]

5. **Expand Both Sides:**
\[ 2304 - 25.6 T_f = 2093 T_f - 41860 \]

6. **Rearrange the Equation:**
Combine like terms:
\[ 2304 + 41860 = 2093 T_f + 25.6 T_f \]
\[ 44164 = 2118.6 T_f \]

7. **Solve for \( T_f \):**
\[ T_f = \frac{44164}{2118.6} \approx 20.84 \, ^\circ C \]

### Final Answer:
The final equilibrium temperature of the lead ball and water mixture is approximately \( 20.84 \, ^\circ C \).

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