Specific latent heat

latent heat of ice to water is 334 kJ/kg

specific heat capacity of water is 4.18X10^3 kJ/kg

250 g water cools down from 80 degree to 0 degree lost energy = (250/1000)(4.18X10^3)(80-0) = 83600 J

Let mass of ice needed be m

heat absorbed for m kg -4 degree C ice to 0 degree water =

(m)(2060)[0-(-4)]+(m)(334X10^3) = 83600

8240m+334000m=83600

342240m = 83600

m=0.2443 kg

m=244.3 g

2014-02-03 13:47:39 補充：
可以的，如果我識的話，會回答你。

2014-02-03 13:49:09 補充：
為什麽你的名稱會變化？

2014-02-04 12:12:41 補充：
4180 和 4200 差不了多小，是有效數字的問題，重點應該在計算的方法，提問者沒有給出數據，幾十年前又用慣了 4.18 J = 1 cal。冇計啦！我其實無教過物理，不知道現在的考試規則。

郭老師，那是因為系統有時會出問題，這個情況之前已出現過多次。

稍後會回復正常。

2014-02-03 15:38:25 補充：
另想問，是否一般用 4200 J/kg/K for shc of water?

2014-02-04 14:41:16 補充：
明白明白～

(｡◕‿◕｡)

唔該兩位不過等我睇一陣先啦如果有唔明係唔係都可以問你地家?

Heat released by water when its temperature falls from 80'C to 0'C
= (250/1000) x 4200 x 80 J = 84000 J
where 4200 J/kg is the specific heat capacity of water

Let m be the mass of ice needed.
Hence, heat absorbed by ice
= m.(2060 x 4 + 334000) J
= 342240m J
(where 334000 J/kg is the latent heat of fusion of water)

By the heat balance equation,
84000 = 342240m
m = 0.245 kg