f.3 Physic

你肯定係2kg ice = =
冇可能嫁bo...應該係水..我估
water gain energy=steam loss energy
咁我地先計steam轉水先放出的energy即係specific latent heat of vaporization
2.26x10^6x0.2=452000J
咁再計water升上100度所需的energy=4200x(100-20)x2=672000J
咁即係話d water足夠令d steam轉晒做水
20℃water gain energy=steam loss latent heat +100℃water loss energy
2x4200x(t-20)=452000+0.2x4200x(100-t)
8400t-168000=452000+84000-840t
8400t+840t=704000
t=76.2℃
加埋就係2.2kg...76.2℃的water

2008-03-09 12:57:35 補充：
d ice一高過0℃就已經開始溶...唔會有20℃冰..
我唔明白2kg ice at 20℃係點

2008-03-09 13:00:47 補充：
要唔要講埋幾幾多% ge水同冰?(我可能唔識 = =)

2008-03-09 13:03:09 補充：
同埋冰ge heat capacity係咩??

2008-03-09 13:15:32 補充：
咁呢..我地首先當唔知 final result product係咩先啦..
咁就計下-20℃ice要升到0℃的water要幾多energy先啦
ice heat capacity+specific latent heat of fusion
=2000x2x(0-(-20))+3.34x10^5x2
=748000J
咁我地再計下100℃steam轉去0℃的water要loss幾多energy啦
latent heat of vaporization+water heat capacity
=452000J+4200x0.2x(100-0)
=536000J
由於748000J>536000J..

2008-03-09 13:21:15 補充：
so我地就知到佢係0℃的ice and water or 0℃以下的ice啦
then..佢要變成0℃以下的冰的energy係:3.34x10^5x2.2=734800J
咁我地一睇就知748000-536000係細過734800J
so就係0℃的ice and water..
咁我地依家轉晒佢係0℃water先...
即係2.2kg0℃water再要loss748000-536000=212000J咁多energy

2008-03-09 13:21:27 補充：
咁212000J可以令幾多mass的0℃water變冰呢??
mass x latent heat of fusion=212000
mass=212000/(10^5x3.34)
=0.63473kg0℃ice啦..
咁0℃ice估0.63473/2.2x100%=28.851%=28.9%啦
咁0℃water佔1-28.9%=71.1%

2008-03-09 13:29:30 補充：
中三用英文讀phy..good~~
我好似冇見過要計% ge 呢d問題
通常都係mc問...仲要淨係問幾多℃的product~
都算f.3最難啦~(but我f.3冇學heat = = 學電..都唔知點解~)