how to calculate the power?

Specific latent heat of vaporization of water:2.26x10^6J/kg
specific heat capacity of water is 4200Jkg^-1℃^-1
the answer tell us steam is produced at a rate of 5gmin^-1
and it is from 25℃ water,so we calculate...
the energy is needed for 5g25℃ water--->5g100℃ water
and the energy is needed for 5g100℃ water-->5g100℃steam
mc△t (△t means the change of temperature) Specific latent heat
=4200x0.005x(100-25) 0.005x2.26x10^6
=1575 11300
=12875J
so we konw that there are required 12875J per min
P=E/t(second)
P=12875/(1x60)
P=214.583333
=215w
the power required to changes 25℃ water to steam is 215w.

2008-04-03 23:06:19 補充：
=1575 11300<---呢度中間漏咗個(加)

2008-04-04 21:52:36 補充：
Gabriella Montez的方法比較正統!!較好~=v=

我個方法有少少唔同, 僅作參考:
Let the mass of water be mkg.
E = ( m )( 4200 )( 100 - 25 ) + ( m )( 2.26 x 10^6 ) = 2575000m
E = Pt
2575000m = P ( m / 0.005 )( 60 )
P = 215W ( 3 s.f. )