In: Chemistry
What is the energy (kcal) that must be removed to take 36.7 gram of water from 100°C steam to 35°C liquid water?
100°C --------------------------> 35°C
(1) (2)
moles of water = 36.7 / 18.01
= 2.04
Q1 = n x delta H vap
= 2.04 x 40.7 x 10^3
= 82937 J
Q2 = m Cp dT
Q2 = 36.7 x 4.184 x (100 -35)
Q2 = 9981 J
Q = Q1 + Q2
Q = 82937 + 9981
Q = 92918 J
Q = 92918 / 4.184 cal
Q = 22208 cal
Q = 22.2 kcal
energy removed = 22.2 kcal