Question
An electric heater which produces 900W of power is used to vaporize water. How much water at 100c can be changed to steam in 3 mins by the heater? (Heat of vaporization=2.26×106j/kg, specific heat capacity of water=4.2×103 j/kg.k).
Answers
if it is already at 100 temp will not change, just boils
2.26*10^6 m = 900 (3 *60)
m = (900*3*60 / 2.26) 10^-6
2.26*10^6 m = 900 (3 *60)
m = (900*3*60 / 2.26) 10^-6
H=Pt
P=900w
t=3min
3*60=180s
H=900*180=162,000J
H=ml
162,000 = m*2.26*10^6
m=162,000/2.26*10^6
=0.072kg
P=900w
t=3min
3*60=180s
H=900*180=162,000J
H=ml
162,000 = m*2.26*10^6
m=162,000/2.26*10^6
=0.072kg
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