An electric kettle takes 10 minutes from 30 Celsius to 100 Celsius. Find the extra time required to vaporize half of the boiling water in the kettle. Assume that no heat is lost to the surroundings.
specific heat capicity of water 4200
specific latent heat of vaporization of water : 2.268 X 10^6
phy既問題一題
2007-10-05 3:05 am
回答 (2)
2007-10-05 3:24 am
✔ 最佳答案
Power P of the electric kettle= E ÷ t
= m c △T ÷ t
= m (4200) (100-30) ÷ (10 × 60)
= 490 m
Energy E' required to vaporize half of the boiling water
= m' l
= (m ÷ 2) 2.268 × 106
= 113400 m
Time required to vaporize half of the boiling water
= E' ÷ P
= (113400 m) ÷ (490 m)
= 2314
So, 2310 s required to vaporize half of the boiling water
2007-10-04 19:29:13 補充：
抱歉! 113400 m 打少左個0應該係 1134000 m不過答案應該啱, 2310 s
2007-10-05 3:25 am
Output of the electric kettle:
(100-30) * 4200 / 10 / 60=490W (J/s)
Time required to vaporize half of the boiling water in the kettle:
2.268 * 10^6 / 490 / 60 / 2 ~= 38.5 minutes
(100-30) * 4200 / 10 / 60=490W (J/s)
Time required to vaporize half of the boiling water in the kettle:
2.268 * 10^6 / 490 / 60 / 2 ~= 38.5 minutes
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