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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 1 year ago
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