急問physics

1kg of ice transform to 1kg of 100 deg C water
energy require = latent heat of fusion + specific heat capicity *100
= 33.44 *10^4 + 100 * 4.17 * 1000
= 751400 J
In order to supply that 751400J, mass of 100 deg C steam need to transform to 100 dec C water
= 751400 / (226.1 * 10^4) kg
= 0.3323 kg
As a result, the mixture is at 100 deg C, where there is 1.3323 kg of 100 deg C water & 0.6677 kg of 100 deg C steam.
混合物是攝氏１００度，其中１．３３２３公斤是水，０．６６７７公斤是水蒸氣。
I hope this can help with your understanding. =)

2007-12-03 21:38:45 補充：
The only equation I use isHeat gain from steam transformed to 100 deg C water = Heat consumed from ice transformed to 0 deg C water then to 100 deg C water

2007-12-03 21:39:17 補充：
Heat consumed from 1kg ice transformed to 0 deg C water then to 100 deg C water= latent heat of fusion * 1kg + specific heat capacity * 1kg *100= 33.44 *10^4 * 1kg + 100 * 4.17 * 1000g= 751400 J

2007-12-03 21:39:46 補充：
Heat gain from 1kg steam transformed to 100 deg C water= latent heat of vapourization * 1kg= 226.1 * 10^4 * 1kg= 2261000 J which is much greater than 751400 JSo only part of steam is transformed to 100 deg C water.

2007-12-03 21:40:37 補充：
Mass of steam transformed to 100 deg C water is= 751400 / 2261000 kg= 0.3323 kgAs a result, the mixture is at 100 deg C, where there is 1.3323 kg of 100 deg C water & 0.6677 kg of 100 deg C steam.

steam + ice = great !