Heat and Enthalpy Solved Problems1. Suppose you mix 21.0 g of water at 52.7° C with 54.9 g of water at 31.5°C in an insulated cup. What is the maximum temperature of the waterafter mixing?Answer:Asystem = -4surroundingsЯhot H,0 — -Ясоol H20Тhot H,0Chot н,047hot H,0 3 -тсоol H20Ссol H,047cоol H,0>(4.184)aTnot H,0 = -(54.9 g) (4.184 )AThot H,0 =-ΔΙ cool H2O°C87.86 )ATrot Hz0 = (229.702) ATeсool Hz0AThot H,0 = (2.614)ATcool H20Tf - 52.7°C = 2.614T; – 31.5°C2.614T; := 21.2°CTf == 8.11°CHeat and Enthalpy Solved Problems2. Use the information provided to determine AH°rxn for the followingreaction3 Fe203 (s) + CO (g) → 2 Fe304 (s) + CO2 (g)ΔΗ ?rxnΔΗ'Fe203 (s)Fe304 (s)CO (g)CO2 (g)-824 kJ/mol-1118 kJ/mol-111 kJ/mol-394 kJ/molAnswer:kJ(-824-molkJ+ -111mol)kJ1118·molkJ+ -394.mol)kJ577mol

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1. Suppose you mix 21.0 g of water at 52.7° C with 54.9 g of water at 31.5° C in an insulated cup. What is the maximum temperature of the water after mixing?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Heat and Enthalpy Solved Problems
1. Suppose you mix 21.0 g of water at 52.7° C with 54.9 g of water at 31.5°
C in an insulated cup. What is the maximum temperature of the water
after mixing?
Answer:
Asystem = -4surroundings
Яhot H,0 — -Ясоol H20
Тhot H,0Chot н,047hot H,0 3 -тсоol H20Ссol H,047cоol H,0
>(4.184)aTnot H,0 = -(54.9 g) (4.184 )
AThot H,0 =-
ΔΙ cool H2O
°C
87.86 )ATrot Hz0 = (229.702) ATe
сool Hz0
AThot H,0 = (2.614)ATcool H20
Tf - 52.7°C = 2.614T; – 31.5°C
2.614T; :
= 21.2°C
Tf =
= 8.11°C
Heat and Enthalpy Solved Problems
2. Use the information provided to determine AH°rxn for the following
reaction
3 Fe203 (s) + CO (g) → 2 Fe304 (s) + CO2 (g)
ΔΗ ?
rxn
ΔΗ'
Fe203 (s)
Fe304 (s)
CO (g)
CO2 (g)
-824 kJ/mol
-1118 kJ/mol
-111 kJ/mol
-394 kJ/mol
Answer:
kJ
(-824-
mol
kJ
+ -111
mol)
kJ
1118·
mol
kJ
+ -394.
mol)
kJ
577
mol

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