How much energy needs to be expended to turn 2 kg water taken at 20 ° C into steam at 100 ° C?

How much energy needs to be expended to turn 2 kg water taken at 20 ° C into steam at 100 ° C? (Specific heat of water 4200 J / kg * ° C, specific heat of vaporization of water 2.3 * 10 to the sixth power J / kg)

Condition:
m = 2kg
t1 = 20 ° C
t2 = 100 ° C
s = 4200 J / kg * ° C
L = 2.3 * 10 ^ 6 J / kg
To find :
E -? J
Decision :
E = Q = Q1 + Q2.
E = Q = c * m * Δt + L * m.
E = Q = c * m * (t2-t1) + L * m.
[E] = J / kg * ° C * kg * ° C + J / kg * kg = J + J = J.
E = Q = 4200 * 2 * 80 + 2.3 * 10 ^ 6 * 2 = 672000 + 4600000 = 5272000 (J).
Answer: 5272 kJ.