Hesting Curve of Weter Evaporation Steam 100- Candensastion 50- Water Melting Freezing lce Jeules The graph above shows the heating curve of water. Use this graph to answer the following question. How much heat is required to heat 28 g of ice at -12.0°C (Celsius) to 28 g of water at 59.0°C? spec. heat of H;0 = 4.184 J/g°C spec. heat of ice = 2.087 J/g°C Heat of Fusion = 6.02 kJ/mol 1 mol of water = 18.0g

Chemistry
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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
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Hesting Curve of Weter
Evaporation
Steam
100-
Condensation
50-
Water
Melting
Freezi ng
Ice
Joules
The graph above shows the heating curve of water. Use this graph to answer the following question.
How much heat is required to heat 28 g of ice at -12.0°C (Celsius) to 28 g of water at 59.0°C?
spec, heat of H20 = 4.184 J/g°C
spec. heat of ice = 2.087 J/g°C
Heat of Fusion = 6.02 kJ/mol
1 mol of water = 18.0g
%3D
%3D
Transcribed Image Text:Hesting Curve of Weter Evaporation Steam 100- Condensation 50- Water Melting Freezi ng Ice Joules The graph above shows the heating curve of water. Use this graph to answer the following question. How much heat is required to heat 28 g of ice at -12.0°C (Celsius) to 28 g of water at 59.0°C? spec, heat of H20 = 4.184 J/g°C spec. heat of ice = 2.087 J/g°C Heat of Fusion = 6.02 kJ/mol 1 mol of water = 18.0g %3D %3D
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