The temperature of a `5mL` of strong acid increases by `5^(@)` when` 5 mL` of strong base is added to it . If `10mL` of each are mixed, temperature should increase by `:`

Options

A. `5^(@)C`
B. `10^(@)C`
C. `15^(@)C`
D. Cannot be known

Explanation

To solve the problem, we need to analyze the situation of mixing a strong acid with a strong base and how the temperature changes as a result of the reaction.

### Step-by-Step Solution:

1. **Understanding the Reaction**: When a strong acid is mixed with a strong base, a neutralization reaction occurs. This reaction is exothermic, meaning it releases heat, which causes the temperature of the solution to increase.

2. **Initial Conditions**: We are given that when 5 mL of strong acid is mixed with 5 mL of strong base, the temperature increases by 5°C. This means that the heat released (Q) during this reaction can be quantified.

3. **Heat of Neutralization**: The heat released during the neutralization can be expressed using the formula:
\[ Q = m \cdot S \cdot \Delta T \]
where:
- \( Q \) = heat released
- \( m \) = mass of the solution (which is proportional to the volume since the density is assumed to be similar)
- \( S \) = specific heat capacity of the solution
- \( \Delta T \) = change in temperature

4. **Doubling the Volume**: Now, if we mix 10 mL of strong acid with 10 mL of strong base, the total volume of the solution becomes 20 mL. The mass of the solution effectively doubles (from 10 mL to 20 mL).

5. **Proportionality of Heat**: Since the heat released during the neutralization is proportional to the volume (or mass) of the reactants, if we double the volume of both the acid and the base, the heat released will also double:
\[ Q_{new} = 2Q_{initial} \]
where \( Q_{initial} \) is the heat released when 5 mL of acid and 5 mL of base are mixed.

6. **Calculating the Change in Temperature**: Since the heat released is doubled, we can express the new change in temperature (\( \Delta T_{new} \)) as:
\[ Q_{new} = m_{new} \cdot S \cdot \Delta T_{new} \]
Given that the specific heat capacity \( S \) remains constant, and the mass \( m_{new} \) is doubled, we can set up the relationship:
\[ 2Q_{initial} = 2m \cdot S \cdot \Delta T_{initial} \]
This implies:
\[ \Delta T_{new} = \Delta T_{initial} \]
Therefore, the increase in temperature remains the same, which is 5°C.

7. **Final Answer**: The temperature increase when 10 mL of strong acid is mixed with 10 mL of strong base will still be 5°C.

### Conclusion:
The temperature should increase by **5°C**.