Answer:
[tex]\boxed{\text{2700 J}}[/tex]
Explanation:
HCl + NaOH ⟶ NaCl + H₂O
There are two energy flows in this reaction.
Heat of reaction + heat to warm water = 0
q₁ + q₂ = 0
q₁ + mCΔT = 0
Data:
m(HCl) = 50 g
m(NaOH) = 50 g
T₁ = 22 °C
T₂ = 28.87 °C
C = 4.18 J·°C⁻¹g⁻¹
Calculations:
m = 50 + 50 = 100 g
ΔT = 28.87 – 22 = 6.9 °C
q₂ = 100 × 4.18 × 6.9 = 2900 J
q₁ + 2900 = 0
q₁ = -2900 J
The negative sign tells us that the reaction produced heat.
The reaction produced [tex]\boxed{\textbf{2900 J}}[/tex].