How much heat is released?

Question

How much heat is released when 75.0 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J/(mol  °C), and the molar heat capacity of ice is 36.4 J/(mol  °C).

Answer 1

Since all of the thermodynamic constants are provided in units of kJ/mol or J/mol, we will first convert 75.0 g of steam / water to moles, and then proceed with the calculations.

moles of steam/water present = 75.0 g x 1 mol / 18 g = 4.17 moles

The calculations will involve several steps:

1). Convert steam at 100º to liquid water @ 100º - phase change, no change in temperature

q = (m)(∆Hvap) = 4.17 moles x 40.67 kJ/mol = 169 kJ

2). Lower temperature of liquid water @ 100º to liquid water @0º

q = mC∆T = (4.17 mol)(75.4 J/molº)(100º) = 31,442 J = 31.4 kJ

3). Convert liquid water @ 0º to solid water @0º - phase change, no change in temperature

q = (m)(∆Hfusion) = (4.17 mols)(6.01 kJ/mol) = 25.1 kJ

4). Lower temperature of ice from 0º to -15º

q = mC∆T = (4.17 mol)(36.4 J/molº)(15º) = 2277 J = 2.28 kJ

Sum all the kJ values to obtain 227.8 kJ of heat is released