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Given the reaction:

[tex]\[ 2K + 2H_2O \rightarrow 2KOH + H_2 \][/tex]

If [tex]\( 19.55 \, \text{g} \)[/tex] of solid [tex]\( K \)[/tex] is added to [tex]\( 500 \, \text{mL} \)[/tex] of water, initially at [tex]\( 20.0^{\circ} \text{C} \)[/tex], and the temperature at the end of the reaction was [tex]\( 61.5^{\circ} \text{C} \)[/tex], what is the heat of solution, [tex]\( q \)[/tex], in joules?

Given:
[tex]\[ C_{\text{soln}} = 4.18 \, \text{J} / \text{g} \cdot ^{\circ}\text{C} \][/tex]
[tex]\[ d_{H_2O} = 1.00 \, \text{g} / \text{mL} \][/tex]


Sagot :

To find the heat of solution, [tex]\( q \)[/tex], let's break the problem down step by step.

1. Determine the Mass of Water:

Given the density of water [tex]\( d_{H_2O} \)[/tex] is [tex]\( 1.00 \, \text{g/mL} \)[/tex], and the volume of water is [tex]\( 500 \, \text{mL} \)[/tex]:
[tex]\[ \text{mass\_water} = 500 \, \text{mL} \times 1.00 \, \text{g/mL} = 500 \, \text{g} \][/tex]

2. Determine the Specific Heat Capacity:

The specific heat capacity [tex]\( C_{\text{soln}} \)[/tex] is given as [tex]\( 4.18 \, \text{J/g}^\circ\text{C} \)[/tex].

3. Calculate the Change in Temperature:

The initial temperature [tex]\( T_{\text{initial}} \)[/tex] is [tex]\( 20.0^\circ\text{C} \)[/tex] and the final temperature [tex]\( T_{\text{final}} \)[/tex] is [tex]\( 61.5^\circ\text{C} \)[/tex]:
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 61.5^\circ\text{C} - 20.0^\circ\text{C} = 41.5^\circ\text{C} \][/tex]

4. Calculate the Heat Absorbed ( [tex]\(q\)[/tex] ):

The formula to calculate the heat absorbed is:
[tex]\[ q = m \times C \times \Delta T \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the water (500 g),
- [tex]\( C \)[/tex] is the specific heat capacity (4.18 J/g°C),
- [tex]\( \Delta T \)[/tex] is the change in temperature (41.5°C).

Substituting the values in, we get:
[tex]\[ q = 500 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 41.5^\circ\text{C} \][/tex]

5. Compute the Heat of Solution:

Performing the multiplication:
[tex]\[ q = 500 \times 4.18 \times 41.5 = 86735 \, \text{J} \][/tex]

Therefore, the heat of solution, [tex]\( q \)[/tex], is [tex]\( 86735 \)[/tex] joules.