PHYSIC HEAT

The latent heat of fusion of ice=3.34 X10^5 J/kg
The specific heat capacity of water =4200J/kgC
The latent heat of vaporization of steam =2.26 X 10^6J/kg
Since the 3rd is much greater than 1 st and 2nd (X100) in value, we can predict that the resullt will be a mixture of steam and boiling water
The absorbed when 0.1 kg of ice is melt
=0.1X3.34X10^6
=3.34X10^5 J
The heat absorbed when the 0 C water become boiling water
=0.1X4200X100
=42000J




2007-12-31 20:38:05 補充：
Sorry, 6th line is 0.1X3.34X10^5=33400JSum of the heat absorbed=33400+42000=75400J----------(*)Heat release when 0.1kg of steam turns to 100 C water =2.26X10^6X0.1=226000J

2007-12-31 20:41:51 補充：
which is greater than the sum to heat absorbed, heat which can be released is in excess (in surplus)The temp is 100C mixture of ice and waterLet m1 be the mass of ateam turns to boiling water75400=m(2.26X10^6)m=0.0334kgwith 0.1334kg of boiling water and 0.0666kg of steam

首先=.=
我自己計完個答案好似有點兒奇怪
我的步驟:
Owing to heat loss = heat gain
Therefore,
specific latent heat of ice and specific heat capacity of melting water=
specific latent heat of steam and specific heat capacity of boiling water
Let the temperature be T,
(0.1)x(3.34x10^5)+(0.1)(4200)(T -0) =(0.1)x(2.26x10^6)+(0.1)(4200)(100-T)
420T+33400=226000+42000-420T
840T=234600
T=279 degree.
=.=自己check 下，錯的話sd番封e-mail俾我，講個答案俾我聽~

0.1x4200x( T - 0 ) = 0.1X4200X( 100 - T )
T-0 = 100-T
2T = 100
T = 50
The mixture temperature is 50*C。