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Heating Curve of Water Steam Evaporation 100- Condensation 50- Water Melting/ Freezing 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

Heating Curve of Water Steam Evaporation 100- Condensation 50- Water Melting/ Freezing 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

Applications and Investigations in Earth Science (9th Edition)
Applications and Investigations in Earth Science (9th Edition)
9th Edition
ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Chapter1: The Study Of Minerals
Section: Chapter Questions
Problem 1LR
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Hesting Curve of Water Steam Evaporation 100- Condensation 50- Water Melting, Freezing 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 H2O = 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
Transcribed Image Text:Hesting Curve of Water Steam Evaporation 100- Condensation 50- Water Melting, Freezing 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 H2O = 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
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