The amount of heat is released during the condensation of 250 g of water vapor at a temperature

The amount of heat is released during the condensation of 250 g of water vapor at a temperature of 100 ° C and cooled to 20 ° C.

Given:
m = 250g = 0.25kg,
L = 2.26MJ / kg,
c = 4200J / (kg * deg),
t1 = 100 ° С,
t2 = 20 ° C;
Find: Q -?
Q = Q1 + Q2,
where
Q1 = L * m – the amount of heat released during steam condensation,
Q2 = m * c * (t1 – t2) – the amount of heat released when the water cools down.
Q = 2.26 * 0.25 MJ + 0.25 * 4200 * 80 J = 649kJ.