When `50 cm^(3)` of a strong acid is added to of an alkali, the temperature rises by `5^(@)C`. If `250 cm^(3)` of each liquid are mixed, the temperature rise would be

Options

A. `5^(@)C`
B. `10^(@)C`
C. `25^(@)C`
D. `20^(@)C`

Answer

To solve the problem, we need to understand the concept of heat of neutralization when mixing a strong acid with a strong base. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
When 50 cm³ of a strong acid is mixed with 50 cm³ of a strong alkali, the temperature rises by 5°C. This indicates that a certain amount of heat is released during the neutralization reaction.

### Step 2: Identify the Heat of Neutralization
For strong acids and strong bases, the heat of neutralization is constant. This means that the amount of heat released per mole of acid-base reaction does not change with the volume of the reactants.

### Step 3: Relate Volume to Temperature Change
In the initial scenario, we have:
- Volume of acid = 50 cm³
- Volume of alkali = 50 cm³
- Temperature rise (ΔT) = 5°C

This temperature rise is due to the heat of neutralization when these two volumes are mixed.

### Step 4: Scale Up the Volumes
Now, if we mix 250 cm³ of the strong acid with 250 cm³ of the strong alkali, we are essentially increasing the volume of both reactants by a factor of 5 (since 250 cm³ is 5 times 50 cm³).

### Step 5: Analyze the Effect on Temperature Change
Since the heat of neutralization is constant and does not depend on the volume of the reactants, the temperature rise (ΔT) will remain the same regardless of the volumes mixed. Therefore, even with the increased volumes, the temperature rise will still be 5°C.

### Conclusion
The temperature rise when 250 cm³ of each liquid is mixed will still be 5°C.

### Final Answer
The temperature rise would be **5°C**.

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