Let's work through this problem step by step.
### Given Data:
1. Mass of solid \( K \) = \( 1.00 \) gram
2. Volume of water = \( 500 \) mL
3. Initial temperature of the water = \( 21.3^\circ C \)
4. Final temperature of the solution after reaction = \( 23.5^\circ C \)
5. Specific heat capacity of the solution, \( c_{\text{soln}} \) = \( 4.18 \text{ J/g}^\circ \text{C} \)
6. Density of the solution, \( d_{\text{soln}} \) = \( 1.00 \text{ g/mL} \)
### Steps for Calculation:
#### Step 1: Calculate the mass of the solution
The mass of the solution can be calculated using the volume of water and the density of the solution.
Since the density \( d_{\text{soln}} \) is \( 1.00 \text{ g/mL} \), the mass of the solution is:
[tex]\[
\text{mass}_{\text{solution}} = \text{volume}_{\text{water}} \times \text{density}_{\text{soln}} = 500.0 \text{ mL} \times 1.0 \text{ g/mL} = 500.0 \text{ grams}
\][/tex]
#### Step 2: Calculate the change in temperature (\(\Delta T\))
The change in temperature (\(\Delta T\)) is given by the final temperature minus the initial temperature.
[tex]\[
\Delta T = 23.5^\circ C - 21.3^\circ C = 2.2^\circ C
\][/tex]
#### Step 3: Calculate the heat absorbed by the solution (\(q\))
The heat absorbed by the solution (\(q\)) can be calculated using the formula:
[tex]\[
q = m \times c_{\text{soln}} \times \Delta T
\][/tex]
where:
- \( m \) is the mass of the solution
- \( c_{\text{soln}} \) is the specific heat capacity of the solution
- \( \Delta T \) is the change in temperature
Substitute the known values into the formula:
[tex]\[
q = 500.0 \text{ grams} \times 4.18 \text{ J/g}^\circ \text{C} \times 2.2^\circ C
\][/tex]
#### Step 4: Calculate the numerical value of heat (\(q\))
[tex]\[
q = 500.0 \times 4.18 \times 2.2 \approx 4598 \text{ J}
\][/tex]
### Conclusion
The heat of reaction, [tex]\( q_{r \times n} \)[/tex], associated with the dissolution and reaction of [tex]\( 1.00 \)[/tex] gram of [tex]\( K \)[/tex] in [tex]\( 500 \)[/tex] mL of water, resulting in the temperature change from [tex]\( 21.3^\circ C \)[/tex] to [tex]\( 23.5^\circ C \)[/tex], is approximately [tex]\( 4598 \text{ J} \)[/tex].
