A solution of `200 mL` of `1 M KOH` is added to `200 mL` of `1 M HCl` and the mixture is well shaken. The rise in temperature `T_(1)` is noted. The experiment is repeated by using `100 mL` of each solution and increase in temperature `T_(2)` is again noted. Which of the following is correct?

Options

A. `T_1 = T_2`
B. `T_2` twice as large as `T_1`
C. `T_1` is twice as large as `T_2`
D. `T_1` is four times as large as `T_2`

Answer

To solve the problem, we need to analyze the heat evolved during the neutralization reaction of KOH and HCl and how it affects the temperature rise in both scenarios.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- KOH (potassium hydroxide) is a strong base, and HCl (hydrochloric acid) is a strong acid. When they react, they produce water and potassium chloride (KCl).
- The reaction is:
\[ \text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O} \]

2. **Heat Evolved in Neutralization**:
- The heat evolved during the neutralization of 1 equivalent of a strong acid with a strong base is approximately 13.7 kcal (or 57.1 kJ).
- Since both KOH and HCl are monobasic, 1 mole of KOH will neutralize 1 mole of HCl.

3. **Calculating Heat for 200 mL of 1 M Solutions**:
- In the first case, we have 200 mL of 1 M KOH and 200 mL of 1 M HCl.
- The total volume is 400 mL, and since both solutions are 1 M, we have 0.2 moles of KOH and 0.2 moles of HCl.
- The heat evolved (q1) can be calculated as:
\[ q_1 = 13.7 \text{ kcal} \times \frac{0.2 \text{ moles}}{1 \text{ mole}} = 13.7 \times 0.2 = 2.74 \text{ kcal} \]

4. **Calculating Heat for 100 mL of 1 M Solutions**:
- In the second case, we have 100 mL of 1 M KOH and 100 mL of 1 M HCl.
- The total volume is 200 mL, and we have 0.1 moles of KOH and 0.1 moles of HCl.
- The heat evolved (q2) can be calculated as:
\[ q_2 = 13.7 \text{ kcal} \times \frac{0.1 \text{ moles}}{1 \text{ mole}} = 13.7 \times 0.1 = 1.37 \text{ kcal} \]

5. **Relating Heat to Temperature Rise**:
- The temperature rise (T) can be calculated using the formula:
\[ q = m \cdot s \cdot \Delta T \]
- Where:
- \( q \) = heat evolved
- \( m \) = mass of the solution (assuming density of water, 1 g/mL, for simplicity)
- \( s \) = specific heat capacity (approximately 1 cal/g°C for water)
- \( \Delta T \) = temperature rise

6. **Calculating Temperature Rise for Each Case**:
- For the first case (200 mL + 200 mL):
\[ 2.74 \text{ kcal} = 400 \text{ g} \cdot 1 \text{ cal/g°C} \cdot T_1 \]
\[ T_1 = \frac{2.74 \times 1000 \text{ cal}}{400 \text{ g}} = 6.85 \text{ °C} \]

- For the second case (100 mL + 100 mL):
\[ 1.37 \text{ kcal} = 200 \text{ g} \cdot 1 \text{ cal/g°C} \cdot T_2 \]
\[ T_2 = \frac{1.37 \times 1000 \text{ cal}}{200 \text{ g}} = 6.85 \text{ °C} \]

7. **Conclusion**:
- Since both temperature rises \( T_1 \) and \( T_2 \) are equal, we conclude that:
\[ T_1 = T_2 \]

### Final Answer:
The correct conclusion is that \( T_1 = T_2 \).