Question

b. Determine the amount of heat release when 465.3g of 100.0oC steam condenses to 21oC liquid water. (100.0oC ΔHvap= 40.6 kJ/mol)

Answer 1

mass of steam m = 465.3 g

molecular weight of water = 18 g/mol

number of moles of water=465.3/18

n=25.85 mol

heat of vaporization= del H=40.6kj/mol

temperature of steam =100 deg c

heat released from steam at 100 deg c to water 100 deg c is = - n * del H

= -25.85*40.6

= -1049.51 kj

heat relesed from water at 100degree to water 21 deg c is = n cp del T

= 25.85 mol * 75.2 j/mol C *(21-100)

= -153,569.68 j

= -153.57 kj

total heat required is = -1049.51 -153.57= -1203.07 kj