Sure! Let's solve this step-by-step:
### Step 1: Identify the given information
- Mass of water (\(m_{\text{water}}\)): 500 grams
- This is because 500 mL of water has a density (\(d\)) of 1.00 g/mL, so the mass is the same as the volume in grams.
- Specific heat capacity of water (\(C_{\text{soln}}\)): 4.18 J/g°C
- Initial temperature (\(T_{\text{initial}}\)): 21.3°C
- Final temperature (\(T_{\text{final}}\)): 23.5°C
### Step 2: Calculate the change in temperature (ΔT)
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \][/tex]
[tex]\[ \Delta T = 23.5\,°C - 21.3\,°C = 2.2\,°C \][/tex]
### Step 3: Calculate the heat absorbed by water (q)
The formula to calculate the heat absorbed by the water is:
[tex]\[ q = m_{\text{water}} \times C_{\text{soln}} \times \Delta T \][/tex]
Substitute the known values into the equation:
[tex]\[ q = 500\,\text{g} \times 4.18\,\text{J/g°C} \times 2.2\,°C \][/tex]
[tex]\[ q = 4598\,\text{J} \][/tex]
### Step 4: Determine the heat of reaction (qrxn)
The heat of reaction (\(q_{\text{rxn}}\)) is equal in magnitude but opposite in sign to the heat absorbed by the water. This is because the heat absorbed by water originates from the reaction.
[tex]\[ q_{\text{rxn}} = -q \][/tex]
[tex]\[ q_{\text{rxn}} = -4598\,\text{J} \][/tex]
Therefore, the heat of reaction [tex]\( q_{\text{rxn}} \)[/tex] is [tex]\(-4598\,\text{J}\)[/tex].
