A solution of `500mL` of `2M KOH` is added to `500mL` of `2M HCl` and the mixture is well shaken. The rise in temperature `T_(1)` is noted. The experiment is then repeated using `250mL` of each solution and rise in temperature `T_(2)` is againg noted. Assume all heat is taken up by the solution `:`

Options

A. `T_(1)=T_(2)`
B. `T_(1)` is 2 times as larger as `T_(2)`
C. `T_(2)` is twice larger as `T_(1)`
D. `T_(1)` is 4 time as larger as `T_(2)`

Explanation

To solve the problem, we need to analyze the heat of neutralization that occurs when KOH is mixed with HCl.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The neutralization reaction between KOH (potassium hydroxide) and HCl (hydrochloric acid) can be represented as:
\[ \text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O} \]
This reaction releases heat, which causes the temperature of the solution to rise.

2. **Calculate Moles of Reactants**:
For the first scenario:
- Volume of KOH = 500 mL = 0.5 L
- Molarity of KOH = 2 M
- Moles of KOH = Molarity × Volume = \(2 \, \text{mol/L} \times 0.5 \, \text{L} = 1 \, \text{mol}\)

Similarly, for HCl:
- Volume of HCl = 500 mL = 0.5 L
- Molarity of HCl = 2 M
- Moles of HCl = \(2 \, \text{mol/L} \times 0.5 \, \text{L} = 1 \, \text{mol}\)

Thus, in the first scenario, we have 1 mole of KOH reacting with 1 mole of HCl.

3. **Heat of Neutralization**:
The heat released during the neutralization is constant for strong acids and bases and can be denoted as \( \Delta H_{neut} \). For simplicity, we can assume it to be a constant value.

4. **Calculate the Temperature Rise (T1)**:
The total heat released (\( q \)) can be expressed as:
\[ q = n \cdot \Delta H_{neut} \]
where \( n \) is the number of moles of the limiting reactant (which is 1 mole in this case).

The heat absorbed by the solution will cause a rise in temperature \( T_1 \):
\[ q = m \cdot c \cdot T_1 \]
where:
- \( m \) = mass of the solution (assuming density = 1 g/mL, for 1000 mL, mass = 1000 g)
- \( c \) = specific heat capacity of water (approximately 4.18 J/g°C)

Thus, we can equate the two expressions for heat:
\[ n \cdot \Delta H_{neut} = m \cdot c \cdot T_1 \]

5. **Repeat with Half Volume**:
In the second scenario, we have:
- Volume of KOH = 250 mL = 0.25 L
- Molarity of KOH = 2 M
- Moles of KOH = \(2 \, \text{mol/L} \times 0.25 \, \text{L} = 0.5 \, \text{mol}\)

Similarly for HCl:
- Volume of HCl = 250 mL = 0.25 L
- Molarity of HCl = 2 M
- Moles of HCl = \(2 \, \text{mol/L} \times 0.25 \, \text{L} = 0.5 \, \text{mol}\)

Again, we have 0.5 moles of KOH reacting with 0.5 moles of HCl.

6. **Calculate the Temperature Rise (T2)**:
The heat released in this case will be:
\[ q = 0.5 \cdot \Delta H_{neut} \]
The mass of the solution will be 500 g (250 mL of solution, assuming density = 1 g/mL).

The equation for heat absorbed becomes:
\[ 0.5 \cdot \Delta H_{neut} = 500 \cdot c \cdot T_2 \]

7. **Relate T1 and T2**:
From the equations derived:
\[ 1 \cdot \Delta H_{neut} = 1000 \cdot c \cdot T_1 \]
\[ 0.5 \cdot \Delta H_{neut} = 500 \cdot c \cdot T_2 \]

Dividing the first equation by the second gives:
\[ \frac{1 \cdot \Delta H_{neut}}{0.5 \cdot \Delta H_{neut}} = \frac{1000 \cdot c \cdot T_1}{500 \cdot c \cdot T_2} \]
Simplifying this leads to:
\[ 2 = 2 \cdot \frac{T_1}{T_2} \]
Hence, we find that:
\[ T_1 = T_2 \]

### Final Answer:
The relationship between the temperature rises is:
\[ T_1 = T_2 \]