Answer is: 100,464 kJ energy are absorbed.
m(water) = 800,0 g.
T₁ = 20°C.
T₂ = 50°C.
ΔT = T₂ - T₁.
ΔT = 50°C - 20°C.
ΔT = 30°C.
c(water) = 4,186 J/g·°C.
Q = m · c · ΔT.
Q = 800 g · 4,186 J/g·°C · 30°C.
Q = 100464 J = 100,464 kJ.
c - specific heat of water.
