蝦餃要用蒸汽烹製。有一蒸鍋可以0.1 kg min-1的速率提供溫度為110度的蒸汽。
(蒸汽 : 比熱容量是2000J kg-1度c-1)
咁我想問如果蒸汽於100度時凝結為水,計算蒸鍋的功率(單位為w)
最好有埋解釋^^
2007-03-28 14:27:35 · 4 個解答 · 發問者 ? 3 in 科學 ➔ 其他:科學
個答案又真係岩wo,
但咁佢比蒸汽 個比熱容量我有咩用,關蒸汽於100度時凝結為水咩事?
2007-03-28 14:51:43 · update #1
個answer係3800w呀--
但我唔係好明
希望會有有心人答我
2007-03-28 15:07:08 · update #2
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.
2007-03-28 19:36:28 · answer #1 · answered by abc 3 · 0⤊ 0⤋
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
2007-03-28 16:30:17 · answer #2 · answered by 天同 7 · 0⤊ 0⤋
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>
2007-03-28 14:59:11 · answer #3 · answered by ? 3 · 0⤊ 0⤋
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)
2007-03-28 14:35:39 · answer #4 · answered by ? 7 · 0⤊ 0⤋