What role doe air density and level above zero play when heating water in an electric kettle?
I have this formula
\(\tau = \frac {c_w m (t_w - t_i)}{\eta P} = \frac {4180 \cdot{2} \cdot{(100-10)}}{0.9 \cdot{ 1200}} \,\mathrm s = 697 \,\mathrm s \dot= 12 \,\mathrm {min}\,. \)
Which is the basic formula to calulate the time it takes to heat 2 kg of water with 90% efficiency and 1200 Watts of power.
No I heard that air density/pressure and level above zero make a difference. Of course in a vaacum boiling is instant, isn't it?
How would those 2 additional variables play into that formula?
Thanks!
+36915 You are only HEATING the water...not BOILING the water....
from your calc it appears as though you are heating the water from 10 to 100 degrees C. The specific heat of water is 4.186 j/(gm-C)
(you are using 4180 j/(kg-C) The specific heat of liquid water is relatively constant below the boiling point.
At a lower pressure (higher altitude) the water will BOIL at a lower temperature....before YOU GET to 100 C, but the enrgy needed to heat each KG of water from it's liquid starting point for each degree C up to the boiling point is still the same.
+36915 No.....the boiling of water (like the melting of ice) occurs at a constant temoerature 100C (lees if the pressure is lowered).....adding more heat will only make it boil faster at 100 C .....you CAN heat up the steam past the boiling point, but not the water.
If you start with very cold ice, you can add heat to warm it up to 0C where it will begin to melt.....adding heat results in faster ice melting, but no temp increase in the ice. You CAN heat up the water past the melting point...up to the boiling point as describd above.
