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)` is 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)`

Explanation

To solve the problem, we need to analyze the two experiments involving the neutralization reaction between KOH (potassium hydroxide) and HCl (hydrochloric acid).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction between KOH and HCl is a neutralization reaction:
\[ \text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O} \]
This reaction releases heat, which causes a rise in temperature.

2. **First Experiment**:
- Volume of KOH solution = 200 mL
- Concentration of KOH = 1 M
- Moles of KOH = Volume (L) × Concentration (mol/L) = \(0.200 \, \text{L} \times 1 \, \text{mol/L} = 0.200 \, \text{mol}\)

- Volume of HCl solution = 200 mL
- Concentration of HCl = 1 M
- Moles of HCl = Volume (L) × Concentration (mol/L) = \(0.200 \, \text{L} \times 1 \, \text{mol/L} = 0.200 \, \text{mol}\)

Since both KOH and HCl are present in equal moles, they will completely neutralize each other.

3. **Total Volume After Mixing**:
The total volume after mixing both solutions = 200 mL + 200 mL = 400 mL.

4. **Temperature Rise in First Experiment**:
The rise in temperature is noted as \(T_1\).

5. **Second Experiment**:
- Volume of KOH solution = 100 mL
- Concentration of KOH = 1 M
- Moles of KOH = \(0.100 \, \text{L} \times 1 \, \text{mol/L} = 0.100 \, \text{mol}\)

- Volume of HCl solution = 100 mL
- Concentration of HCl = 1 M
- Moles of HCl = \(0.100 \, \text{L} \times 1 \, \text{mol/L} = 0.100 \, \text{mol}\)

Again, both KOH and HCl are present in equal moles and will completely neutralize each other.

6. **Total Volume After Mixing in Second Experiment**:
The total volume after mixing both solutions = 100 mL + 100 mL = 200 mL.

7. **Temperature Rise in Second Experiment**:
The rise in temperature is noted as \(T_2\).

8. **Comparing the Two Experiments**:
In both experiments, the number of moles of KOH and HCl is the same, which means the amount of heat released during the neutralization is the same. The only difference is the total volume of the solution, which affects the temperature rise.

However, since the heat released is the same and the total volume is different, we can conclude that the temperature rise will be the same in both cases because the concentration of the reactants remains the same.

9. **Conclusion**:
Therefore, we can conclude that \(T_1 = T_2\).

### Final Answer:
The correct statement is that the rise in temperature \(T_1\) is equal to the rise in temperature \(T_2\).