The specific heat of lead is `0.030 cal//(g^(@)C)`. 300 g of lead shot at `100^(@)C` is mixed with 100g of water at `70^(@)C` in an insulated container. The final temperature of the mixture is

Options

A. `100^(@)C`
B. `85.5^(@)C`
C. `79.5^(@)C`
D. `72.5^(@)C`

Answer

To solve the problem, we will use the principle of calorimetry, which states that the heat lost by the hot substance (lead) will be equal to the heat gained by the cold substance (water) in an insulated system.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of lead (m_lead) = 300 g
- Specific heat of lead (c_lead) = 0.030 cal/(g°C)
- Initial temperature of lead (T_initial_lead) = 100°C
- Mass of water (m_water) = 100 g
- Specific heat of water (c_water) = 1 cal/(g°C)
- Initial temperature of water (T_initial_water) = 70°C

2. **Set up the heat loss and gain equations:**
- Heat lost by lead = Heat gained by water
- For lead: \( Q_{lead} = m_{lead} \cdot c_{lead} \cdot (T_{initial\_lead} - T_f) \)
- For water: \( Q_{water} = m_{water} \cdot c_{water} \cdot (T_f - T_{initial\_water}) \)

3. **Substitute the known values into the equations:**
- Heat lost by lead:
\[ Q_{lead} = 300 \cdot 0.030 \cdot (100 - T_f) \]
- Heat gained by water:
\[ Q_{water} = 100 \cdot 1 \cdot (T_f - 70) \]

4. **Set the heat lost equal to the heat gained:**
\[ 300 \cdot 0.030 \cdot (100 - T_f) = 100 \cdot (T_f - 70) \]

5. **Simplify the equation:**
\[ 9 \cdot (100 - T_f) = T_f - 70 \]
\[ 900 - 9T_f = T_f - 70 \]

6. **Rearranging the equation:**
\[ 900 + 70 = T_f + 9T_f \]
\[ 970 = 10T_f \]

7. **Solve for \( T_f \):**
\[ T_f = \frac{970}{10} = 97°C \]

8. **Final temperature of the mixture:**
\[ T_f = 79.0°C \]

### Final Answer:
The final temperature of the mixture is approximately **79.0°C**.