300 g of water, the initial temperature of which was 400C, cooled down to 20C. How much heat was released during this?

300 g of water, the initial temperature of which was 400C, cooled down to 20C. How much heat was released during this? (specific heat capacity of water 4200 J / (kg C))

Given:
m = 300g = 0.3 kg
t1 = 400 ° C
t2 = 200 ° C
s = 4200 J / kg * ° C
Q-? J
The amount of heat Q is found by the formula Q = c * m * (t1-t2)
Q = 4200 J / kg * ° C * 0.3 kg * (400 ° C-200 ° C) = 252 000 J = 252 kJ