Question
How much heat must be added to 20 grams of water at 100 degrees Celsius to convert it to steam?
My work
Q = (20)(4.18 x 10^3)(100)
= 8360000 J
Is that the correct answer?
My work
Q = (20)(4.18 x 10^3)(100)
= 8360000 J
Is that the correct answer?
Answers
No, it is already at 100, the boiling point
You want
.020 kg * Heat of Vaporization of water in Joules/ kg
so
.020 kg * 2.26*10^6 J/kg
You want
.020 kg * Heat of Vaporization of water in Joules/ kg
so
.020 kg * 2.26*10^6 J/kg
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