Answer:
302 kj heat is released by lowering the temperature
1680kJ
Heat 500 g ice from -20 to 0 deg C:
500 g x 2.03 j/g C x 20 deg = 20300 J = 20.3 kJ (20 kJ)
Convert 500 g ice at 0 deg to 500 g water at 0 deg:
500 g / 18.015 g/mol = 27.75 mol
27.75 mol x 6.02 kJ/mol = 167.0 kJ (167 kJ)
Heat 500 g water from 0 deg to 100 deg C:
500 g x 4.2 J/g C x 100 deg = 210000 J = 210 kJ (210 kJ)
Convert 500 g water at 100 deg to 500 g steam at 100 deg C:
27.75 mol x 40.7 kJ/mol = 1129.4 kJ (1130 kJ)
Heat 500 g steam from 100 deg to 250 deg C:
500 g x 2.0 J/g C x 150 deg = 50000 J = 50 kJ (50 kJ)
Add all the steps:
20 + 167 + 210 + 1130 + 50 = 1577 kJ (1580 kJ to 3 sig. figures)
Are you certain the book says 1680 and not 1580 kJ?
EDIT: Math error in heating steam; should be 150 kJ not 50 kJ
ANSWER: 20 + 167 + 210 + 1130 +150 = 1677 kJ
Using 3 significant figures, the answer is 1680 kJ
Answer: Ecell = -0.110volt
Explanation:
Zn--->Zn^+2 + 2e^-.........(1) oxidation
Cu^2+ 2e^- --->Cu........(2)reduction
Zn + Cu^2+ ----> Cu + Zn^+2 (overall
For an electrochemical cell, the reduction potential set up is given by
E(cell) = E(cathode) - E(anode)
E(cell) = E(oxidation) - E(reduction)
E(cathode) = E(oxidation)
E(anode) = E(reduction)
Given that
E(oxidation) = -0.763v
E(reduction) = +0.337v
E(cell) = -0.763 - (+0.337)
E(cell) = -0.763- 0.337
E(cell) = -0.110volt
I would be neutral because h20 has a impact on water