When 20 ml of `0.1 M HCl` is mixed with 20 ml of `0.1 M NaOH`, the rise in temperature is `T_(1)`. When the experiment is repeated using 20 ml of `0.2 M HCl` and `0.2 M NaOH` solutions, the rise in temperature is `T_(2)`. Then

Options

A. `T_(1)=T_(2)`
B. `T_(1)=2T_(2)`
C. `T_(1)=4T_(2)`
D. `2T_(1)=T_(2)`

Explanation

To solve the problem, we need to analyze the two scenarios involving the mixing of hydrochloric acid (HCl) and sodium hydroxide (NaOH) solutions and their effect on temperature change.

### Step-by-Step Solution:

1. **Identify the Moles of HCl and NaOH in Each Case:**
- For the first case (0.1 M solutions):
- Volume of HCl = 20 ml = 0.020 L
- Molarity of HCl = 0.1 M
- Moles of HCl = Molarity × Volume = 0.1 mol/L × 0.020 L = 0.002 moles

- Volume of NaOH = 20 ml = 0.020 L
- Molarity of NaOH = 0.1 M
- Moles of NaOH = 0.1 mol/L × 0.020 L = 0.002 moles

2. **Calculate the Concentration After Mixing:**
- Total volume after mixing = 20 ml + 20 ml = 40 ml = 0.040 L
- Concentration of HCl after mixing:
\[ \text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Total Volume}} = \frac{0.002 \text{ moles}}{0.040 \text{ L}} = 0.050 \text{ M} \]
- Concentration of NaOH after mixing:
\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Total Volume}} = \frac{0.002 \text{ moles}}{0.040 \text{ L}} = 0.050 \text{ M} \]

3. **Repeat the Process for the Second Case (0.2 M Solutions):**
- For the second case:
- Volume of HCl = 20 ml = 0.020 L
- Molarity of HCl = 0.2 M
- Moles of HCl = 0.2 mol/L × 0.020 L = 0.004 moles

- Volume of NaOH = 20 ml = 0.020 L
- Molarity of NaOH = 0.2 M
- Moles of NaOH = 0.2 mol/L × 0.020 L = 0.004 moles

4. **Calculate the Concentration After Mixing for the Second Case:**
- Total volume after mixing = 20 ml + 20 ml = 40 ml = 0.040 L
- Concentration of HCl after mixing:
\[ \text{Concentration of HCl} = \frac{0.004 \text{ moles}}{0.040 \text{ L}} = 0.100 \text{ M} \]
- Concentration of NaOH after mixing:
\[ \text{Concentration of NaOH} = \frac{0.004 \text{ moles}}{0.040 \text{ L}} = 0.100 \text{ M} \]

5. **Relate the Temperature Changes:**
- In the first case, the temperature rise is \( T_1 \) with a concentration of 0.050 M.
- In the second case, the temperature rise is \( T_2 \) with a concentration of 0.100 M.
- The concentration in the second case is double that of the first case. Since the heat of neutralization is proportional to the concentration of the reactants, we can conclude:
\[ T_2 = 2 \times T_1 \]

### Final Conclusion:
Thus, the relationship between the temperature rises is \( T_2 = 2 T_1 \).