Answer:
611.064 kJ
Explanation:
Given :
m = 200 mL = 200 g
Specific heat of ice = 2.06 J/g°C
Q = mcΔt
Δt = 0 - (-22) = 22
Q1 = 200 * 2.06 * 22 = 9064 J
Q2 = Melt 0 °C solid ice into 0 °C liquid water:
Q2 = m · ΔHf ; ΔHf = heat of fusion of water = 334j/g
Q2 = 200 * 334 = 66800 J
Q3 : Heat to convert from 0°C to 100°C
Q3 = mcΔt ; c = 4.19 J/g°C ; Δt = (100 - 0) = 100
Q3 = 200 * 4.19 * 100 = 83800 J
Q4: heat required to boil water to steam
Q = m · ΔHv
Hv = heat of vaporization of water = 2257 J/g
Q4 = 200 * 2257 = 451400 J
Total Q = Q1 + Q2 + Q3 + Q4
Q = 9064 + 66800 + 83800 + 451400
Q = 611,064 Joules
Q = 611.064 kJ
