Final answer:
To find the total heat evolved when converting 1.00 mol of steam at 150.0 °C to ice at -50.0 °C, we perform multiple calculations using specified heat capacities, molar heat of vaporization, and molar heat of fusion, then sum the heat released at each stage.
Step-by-step explanation:
To calculate how much heat is evolved in converting 1.00 mol of steam at 150.0 °C to ice at -50.0 °C, we must consider the heat required for cooling the steam to 100 °C, condensing the steam to water at 100 °C, cooling the water from 100 °C to 0 °C, freezing the water at 0 °C to ice, and finally cooling the ice from 0 °C to -50.0 °C.
The heat capacity of steam (2.01 J/(g·°C)) and of ice (2.09 J/(g·°C)) are given, along with the molar heat of vaporization of water (40.7 kJ/mol at 100°C) and the molar heat of fusion of water (6.01 kJ/mol at 0°C). Additionally, the molecular weight of water (H₂O) is approximately 18.015 g/mol.
To solve the problem, we must conduct a multi-step calculation using the given heat capacities, molar heat of vaporization, and molar heat of fusion:
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- Calculate the heat released by cooling the steam from 150.0 °C to 100.0 °C.
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- Calculate the heat released by condensing the steam at 100.0 °C to liquid water.
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- Calculate the heat released by cooling the water from 100.0 °C to 0.0 °C.
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- Calculate the heat released by freezing the water at 0 °C to ice.
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- Calculate the heat released by cooling the ice from 0 °C to -50.0 °C.
The sum of these values will give the total heat evolved in the process.