2) Suppose that 29.02 g of ice at -12.8°C is placed in 60.63 g of water at 91.7°

Question

2) Suppose that 29.02 g of ice at -12.8°C is placed in 60.63 g of water at 91.7°C in a perfectly insulated vessel. Calculate the final temperature. (The molar heat capacity for ice is 37.5 J K-1 mol-1 and that for liquid water is 75.3 J K-1 mol-1. The molar enthalpy of fusion for ice is 6.01 kJ/mol. You must answer in Kelvin, not °C.)

3A) A 1.51 g sample of caffeine (C8H10N4O2) burns in a constant-volume calorimeter that has a heat capacity of 7.79 kJ/K. The temperature increases from 297.65 K to 302.58 K. Determine the heat (qv) associated with this reaction.

3B) Now use the data above to find E for the combustion of one mole of caffeine.

Explanation / Answer

2) The heat energy gained by ice = heat energy lost by water

The heat energy gained by ice =no.of moles of ice changing from -12.8C to 0Cx latent heat of fusion of ice+no.of moles of ice x specific heat capacity of icex deltaT

                                             = 29.02g/18g.mol-1 x 6.01kJ/mol +29.02g/18g.mol-1 x 37.5 J/mol K x [T-273]K

heat energy lost by water = moles of water x molar heat capacity of water x deltaT

                                      = 60.63g/18g mol-1 x 75.3 J/mol.K x [364.7-T]K

Hence:

29.02g/18g.mol-1 x 6.01kJ/mol +29.02g/18g.mol-1 x 37.5 J/mol K x [T-273]K = 60.63g/18g mol-1 x 75.3 J/mol.K x [364.7-T]K

by simplifying:

T = 316.2K (THIS IS THE FINAL TEMPERATURE OF THE MIXTURE)

3A) Q = heat capacity x T
Q = 7.79kJ/K x 4.93 K = 38.4kJ

3B) E for the combustion of one mole of caffeine:At constant volume the amount of heat energy of the system= change in internal energy of the system

38.4kJ / (1.51g / 194g/mole) = 4935.73 kJ/mole

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

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