Answer :
Given:
0.103 grams of Zinc (solid)
50 ml of HCl
Density of HCl = 1.0 g/mL
Initial Temperature (Ti) = 22.5 C
Final Temperature (Tf) = 23.7 C
Solve for the mass of HCl:
50 mL * 1.0 g/mL = 50 g
Assume that the Cp for HCl is similar to the Cp of water:
q = mCpdT
= mCp (Tf - Ti)
= 50 g * 4.18 J/gC * (23.7 - 22.5)
q = 250.8 J = H
0.103 grams of Zinc (solid)
50 ml of HCl
Density of HCl = 1.0 g/mL
Initial Temperature (Ti) = 22.5 C
Final Temperature (Tf) = 23.7 C
Solve for the mass of HCl:
50 mL * 1.0 g/mL = 50 g
Assume that the Cp for HCl is similar to the Cp of water:
q = mCpdT
= mCp (Tf - Ti)
= 50 g * 4.18 J/gC * (23.7 - 22.5)
q = 250.8 J = H