Question
In: Chemistry
What is the energy (kcal) that must be removed to take 36.7 gram of water from...
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?
Solutions
Expert Solution
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
queen_honey_blossom answered 2 years agoRelated Solutions
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