Answer: 21,072.24KJ
Explanation:
increasing the temperature of 420 kg of water from 25 degrees C to 37degrees C. The change in temperature is ∆T = (37– 25) degrees C = 12 degrees C.
The specific heat capacity of water is 4,181 J / kg degree C, so the equation gives:
Q=mC ∆T
Q = 420 kg × 4181 J / kg degree C × 12 degrees C
=21, 072,240 J = 21,072.24 kJ