how many calories would liberated when one mole of steam at 100c condenses to H2O at 100c?
To calculate the amount of calories liberated when one mole of steam at 100 degrees Celsius condenses to water (H2O) at 100 degrees Celsius, we need to use the enthalpy of vaporization and the heat capacity of water.
1. Find the enthalpy of vaporization: The enthalpy of vaporization (ΔH_vap) for water is given as 40.7 kJ/mol. This value represents the amount of energy required to convert one mole of water from its liquid state to its gaseous state at its boiling point.
2. Convert kJ/mol to calories/mol: We need to convert the units from kilojoules per mole to calories per mole. Since there are 4.184 joules in a calorie, we can multiply the enthalpy of vaporization by 4.184 to convert it to calories per mole. Therefore, the enthalpy of vaporization is 40.7 kJ/mol x 4.184 cal/J = 170.05 cal/mol.
3. Calculate the heat released when one mole of steam condenses: When one mole of steam at 100 degrees Celsius condenses to water at 100 degrees Celsius, it releases the same amount of heat as the enthalpy of vaporization. Therefore, the number of calories liberated is equal to the enthalpy of vaporization, which is 170.05 cal/mol.
So, one mole of steam at 100 degrees Celsius condensing to water at 100 degrees Celsius would liberate approximately 170.05 calories.