Answer :
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.
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.