0.5000 moles of steam at 200°C was chilled in a ultra-low temperature freezer to -90°C. If the heat of vaporization of steam is 540. cal/gram, the heat of fusion of water is 80.0 cal/gram, the specific heat of steam is 0.500 cal/gram-°C, the specific heat of water is 1.000 cal/gram-°C, and the specific heat of ice is 0.500 cal/gram-°C, how many kJ of energy is released, assuming no energy loss to the environment? 1 calorie = 4.184 Joules.
0.5000 moles of steam at 200°C was chilled in a ultra-low temperature freezer to -90°C. If the heat of vaporization of steam is 540. cal/gram, the heat of fusion of water is 80.0 cal/gram, the specific heat of steam is 0.500 cal/gram-°C, the specific heat of water is 1.000 cal/gram-°C, and the specific heat of ice is 0.500 cal/gram-°C, how many kJ of energy is released, assuming no energy loss to the environment? 1 calorie = 4.184 Joules.
Chemistry by OpenStax (2015-05-04)
1st Edition
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter10: Liquids And Solids
Section: Chapter Questions
Problem 44E: The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why.
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0.5000 moles of steam at 200°C was chilled in a ultra-low temperature freezer to -90°C. If the heat of vaporization of steam is 540. cal/gram, the heat of fusion of water is 80.0 cal/gram, the specific heat of steam is 0.500 cal/gram-°C, the specific heat of water is 1.000 cal/gram-°C, and the specific heat of ice is 0.500 cal/gram-°C, how many kJ of energy is released, assuming no energy loss to the environment? 1 calorie = 4.184 Joules.
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