madey21
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[tex]$100.0 \, \text{mL}$[/tex] of [tex]$0.50 \, \text{M} \, \text{Ba(OH)}_2$[/tex] is added to [tex]$100.0 \, \text{mL}$[/tex] of [tex]$1.00 \, \text{M} \, \text{HCl}$[/tex] in a coffee cup calorimeter at [tex]$20.0^{\circ} \text{C}$[/tex]. The final temperature was [tex]$27.0^{\circ} \text{C}$[/tex].

Given:
[tex]\[
C_{\text{soln}} = 4.2 \, \text{J/g}^{\circ} \text{C} \quad d_{\text{soln}} = 1.00 \, \text{g/mL}
\][/tex]

What is the heat of solution, [tex]$q$[/tex], in joules?


Sagot :

Sure, let's walk through the detailed steps to find the heat of solution [tex]\( q \)[/tex]:

### 1. Determine the total volume of the solution:
Given:
- Volume of [tex]\( \text{Ba(OH)}_2 \)[/tex] solution = 100.0 mL
- Volume of [tex]\( \text{HCl} \)[/tex] solution = 100.0 mL

Total volume of the solution:
[tex]\[ 100.0 \, \text{mL} + 100.0 \, \text{mL} = 200.0 \, \text{mL} \][/tex]

### 2. Calculate the mass of the solution:
Given:
- Density of the solution, [tex]\( d_{\text{soln}} = 1.00 \, \text{g/mL} \)[/tex]
- Since density is mass per unit volume, the mass of the solution can be found by:
[tex]\[ \text{Mass of the solution} = \text{Volume} \times \text{Density} \][/tex]

Thus:
[tex]\[ \text{Mass of the solution} = 200.0 \, \text{mL} \times 1.00 \, \text{g/mL} = 200.0 \, \text{g} \][/tex]

### 3. Determine the change in temperature:
Given:
- Initial temperature, [tex]\( T_{\text{initial}} = 20.0^\circ \text{C} \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 27.0^\circ \text{C} \)[/tex]

Change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 27.0^\circ \text{C} - 20.0^\circ \text{C} = 7.0^\circ \text{C} \][/tex]

### 4. Calculate the heat of solution [tex]\( q \)[/tex]:
Given:
- Specific heat capacity of the solution, [tex]\( C_{\text{soln}} = 4.2 \, \text{J/g} \cdot^\circ \text{C} \)[/tex]

Heat of solution ([tex]\( q \)[/tex]) is calculated using the formula:
[tex]\[ q = \text{mass} \times \text{specific heat capacity} \times \Delta T \][/tex]

Substitute the values:
[tex]\[ q = 200.0 \, \text{g} \times 4.2 \, \text{J/g} \cdot^\circ \text{C} \times 7.0^\circ \text{C} \][/tex]

So:
[tex]\[ q = 200.0 \times 4.2 \times 7.0 = 5880.0 \, \text{J} \][/tex]

### Final Answer:
The heat of solution, [tex]\( q \)[/tex], is [tex]\( 5880.0 \, \text{J} \)[/tex].