唔係好明呢題ｕｐ乜？教下我（中三ｐｈｙｓ）

0.1kg 的100度水變為100度的汽要汽化熱:
0.1x2260000J=226000J
0.1kg的100度汽變為110度的汽要汽化熱:
0.1x(110-100)x2000=2000J
由室溫升至100度的水需熱相對較小,忽略不計
故1min內需熱(即蒸汽鍋供熱)為:
226000J+2000J=228000J
蒸汽鍋功率=228000J/60sec=3800watt

2007-03-29 22:39:08 補充：
本人為此問題的回答者,第三行的"熱量"誤寫為"汽化熱",謹此致欠.望問者注意.

The power supplied by the steaming pot (蒸鍋) is used to vapoourize the water at 100 degC (this is given) and to raise its temperature by 10 deg.C from 100 to 110 deg.C.

Since power is the rate of energy per unit time, thus take a unit time of 1 second,
the nass of steam produced = 0.1/60 kg/s

Heat required to turn it to steam = (0.1/60)x(2.26x10^6) J/s
[where 2.26x10^6 J/kg is the latent heat of vapourization of water]

Heat required to raise the steam temperature from 100-110 deg.C
= (0.1/60)x(2000)x(110-100) J/s

Hence, power needed = (0.1/60)x(2.26x10^6) + (0.1/60)x(2000)x(110-100) J/s
= 3800 J/s = 3800 W

我唔知條題目系咪俾少左野@@"
據我理解,由於蒸鍋可將100度水升溫並將其氣化至110度的蒸氣,
運用 "能量=質量x比熱容量x溫度差":
設有0.1kg的水(即設質量為 0.1kg)
能量=0.1x2000x(110-100)
= 2000J
據"功率=能量/加熱所用時間(秒)"
功率=2000/60
=33.3W
唔知岩唔岩,如有錯漏,唔好見怪>

佢無比時間你，
你可以自己set時間。

設時間 = 1 min
功率
= 能量 / 時間
= ( 0.1 x 2.26 x 10^ 6 ) J / 1 min
=226000J / 60 s
= 3766 W
= 3800W (取2位有效)