Physic heat question

specific heat capacity = 4200 J/kg C
latent heat of vapourization of water = 2.26x10^6 J/kg

energy provided by the heater:
P = E/t
E = (P)(t)
= (1x1000)(10x60)
= 600000J

Energy needed to rasie the water from 20 degree celsius to 100 degree celsius :
E = (0.5)(4200)(100-20)
= 168000 J

Energy needed to vapourize m kg of water:
E = mlv
= m(2.26x10^6)

(2.26x10^6m)+(168000) = 600000
m = 0.191
So, the answer is A

specific heat capacity is 4200J/kg C?
not specific heat capacity of water is 4200J/kg C?

E=mc*T + ml
E=Pt
So, Pt=mc*T + ml
In this question,
P=1000 W
t=10 X 60 s
*T=100-20
m= 0.5 kg
C= 4200
l = 2.26X10^6

After substituting all the variables in the above equation,
0.5(4200)(100-20) + m(2.26 X 10^6) = 1000 (10 X 60)
the answer is A