Help with caliometer problems 1. The following data were obtained when measuring

Question

Help with caliometer problems 1. The following data were obtained when measuring the heat capacity of a calorimeter: Volume of cold water = 100 mL Initial temp. of cold water = 22.50 degree C Volume of hot water = 100 mL Initial temp. of hot water = 42.35 degree C Final temp. of water = 31.80 degree C Follow the procedure in Part 1 of the Calculations to determine the heat capacity of the calorimeter (Cp). 2. The following data were obtained when measuring delta H neutralization for the reaction of HCl with NaOH. Volume of HCl = 100 mL Molarity of HCl = 2.00 M Volume of NaOH = 100 mL Molarity of NaOH = 2.00 M Average initial temperature of the acid and base = 22.50 degree C Final temperature = 35.50 degree C heat capacity of the calorimeter (Cp) = 50 J/K Follow the procedure in Part 2 of the Calculations to determine the molar enthalpy of neutralization (delta H neutralization).

Explanation / Answer

1. qlost by hot water = q gained by hot water

qlost by hot water = m*S*DT

mass of hot water = 100 ml*1 g/ml = 100 grams

S = specific heat of water = 4.18 j/g.c

DT = 35.5-22.5 = 13 c

qlost by hot water = 100*4.18*13 = 5434 joule. = 5.434 kj

heat capacity of calorimeter = 5.434/13 = 0.418 kj/c

2

q released = m*S*DT

mass of solution = 200*1 = 200 grams

S = specific heat of solution= 4.18 j/g.c

DT = 35.5-22.5 = 13 C

q released = m*s*DT+Cp*DT

= 200*4.18*13 + 50*13

q = 11.518 Kj

no of mole of HCl = 0.1*2 = 0.2 mol

q = -DH

DH = -11.518/0.2 = -57.6 kj/mol

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.