First, we will get the heat required to raise the 500 g of water from 50°C to 100°C = m*C*ΔT
when m is the mass = 500 g
and C is the specific heat capacity of water = 4.19
ΔT change in temperature = 50 °C
by substitution:
q = 0.5 Kg * 4.19 * 50°C
=104.75 KJ
∴ heat left to boil the water= 500KJ - 104.75KJ = 395 KJ
the heat required to boil water from 100°C to steam = mass *latent heat of vaporization
395KJ = M * 2.26 x 10^3
Mass = 0.17Kg = 170 g
∴ water remain= 500 g - 170 = 330 g