Physics- Latent heat

Let the mass of ice cube be m1 kg and that of steam be m2 kg.

Assume there is no energy lost to the surroundings.

By the conversation of energy,

Energy gained = Energy lost

m1(3.34*10^5) + m1(4200)(50 - 0) = m2(2.26*10^6) + m2(4200)(100 - 50)

544000m1 = 2470000m2

m1 / m2 = 2470000 / 544000

m1 : m2 = 4.540441176 : 1

= 4.54 : 1 (3 sig.fig)

Therefore, the ratio of the mass of ice to the mass of steam is 4.54 : 1

assume that no heat is released to the surrounding
Let the mass of ice be m, mass of steam be a
m*3.34*10^5+4200*50*m=a*2.26*10^6+4200*50*a
m*3.34*10^5=a*2.26*10^6
m/a=2.26*10^6/3.34*10^5
=6.77
so m:a=6.77

2009-12-22 10:44:01 補充：
sorry I am wrong. a not equals to m

2009-12-22 10:44:19 補充：
chiyuen0622 is right