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標題: principle of mixing + specific latent heat [打印本頁]

作者: PA 時間: 15/12/2007 03:48 PM
標題: principle of mixing + specific latent heat
Where necessary, take specific heat capacity of water = 4200 J kg^-1 ℃^-1 , specific heat capacity of ice = 2100 J kg^-1 ℃^-1, specific latent heat of fushion of ice = 3.34*10^5 Jkg^-1 and specific latent heat of vaporization of water 2.26*10^6 Jkg^-1

1. Determine the final temperature of a mixture of 5g steam at 100℃ with 5g ice at 0℃.
A. 50℃
B. 66.6℃
C. 85.2℃
D. 100℃

答案是D，但為何是100℃？如果計？又不知steam的specific heat capacity...

[ 本帖最後由木守宮的蛙糧-PA 於 16/12/2007 09:23 AM 編輯 ]

作者: fish 時間: 15/12/2007 04:13 PM

原帖由 木守宮的蛙糧-PA 於 15/12/2007 03:48 PM 發表

 登錄/註冊後可看大圖

Where necessary,take specific heat capacity of water = 4200 J kg^-1℃^-1 , specific heat capacity of ice = 2100 J kg^-1 ℃^-1, specificlatent heat of fushion of ice = 3.34*10^5 Jkg^-1 and specific latentheat of vaporization of water 2.26*10^6 Jkg^-1

1. Determine the final temperature of a mixture of 5g steam at 100℃ with 5g ice at 0℃.
A. 50℃
B. 66.6℃
C. 85.2℃
D. 100℃

不一定要用掉所有data的
Ice已在0℃, steam已在100℃....兩者的specific heat capacity在這情況下不用了~
做MC嘛....不用甚麼長step了
(1)Energy needed to melt the ice = (5/1000kg) * (3.34*10^5 Jkg^-1) = 1670J
(2)Energy need to heat melted ice from 0℃ to 100℃ = (5/1000kg) * (4200J kg^-1℃^-1) * (100℃) = 2100J

(3)Energy involved to condense the steam to water = (5/1000kg) * (2.26*10^6Jkg^-1) =11300J

(1)+(2) = 3770J
3770J 比11300J 細，所以當兩者的heat transfer完後，仍是100℃
偶Phy不太好啦.....這樣解你看得明嗎@@?

edit:
加回unit及加上bracket會好看些吧@@

作者: PA 時間: 16/12/2007 09:23 AM
Where necessary, take specific heat capacity of water = 4200 Jkg^-1℃^-1 , specific heat capacity of ice = 2100 J kg^-1 ℃^-1,specificlatent heat of fushion of ice = 3.34*10^5 Jkg^-1 and specificlatentheat of vaporization of water 2.26*10^6 Jkg^-1

10. A 100W immersion heater needs 200s to melt completely 0.6kg of ice at 0℃.
(a) Find the specific latent heat of fusion of ice.

答案是3.33*10^3 Jkg^-1，但我計到3333333 1/3，會是答案錯嗎?

作者: sapphire 時間: 16/12/2007 10:13 AM
energy provided by the heater = Pt = 100 * 200 = 20000 J

20000 = mL
20000 = 0.6L
L = 33333.33 J kg^(-1)

the question is a little bit strange......
if the question asks u to find some constant which is well defined already, (e.g. specific latent heat of ice, radius of earth, mass of sun......), the ans should be very closed to that value. However, this ans is far away from that value......
original value of specific latent heat of ice = 3.34 * 10^5 J kg^(-1)
ans = 33333.33 J kg^(-1)

[ 本帖最後由 sapphire 於 16/12/2007 10:27 AM 編輯 ]

作者: PA 時間: 16/12/2007 12:51 PM

原帖由 sapphire 於 16/12/2007 10:13 AM 發表

 登錄/註冊後可看大圖

energy provided by the heater = Pt = 100 * 200 = 20000 J

20000 = mL
20000 = 0.6L
L = 33333.33 J kg^(-1)

the question is a little bit strange......
if the question asks u to find some constant ...

呃...原來power打少了個0...
10. A1000Wimmersion heater needs 200s to melt completely 0.6kg of ice at 0℃.
(a) Find the specific latent heat of fusion of ice.
Pt=ml
1000*200=0.6l
200000=0.6l
l=200000/0.6
l=333333.3333
不過不是3.33*10^3
真怪的...

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