Question

calculate the energy by kj to heat 20000 liters of water from 35 to 80 °C

Answer 1

Energy Needed = (specific heat of water) × (mass[g]) × (temperature [K(℃is also okay)])
In this case, E=4.217[J/g・K]×1000[g/l] × 20000[l] × (80-35)[K] =3795300kJ

Answer 2

Q = m•C•ΔT
Q= quantity of heat needed
m=mass in grams
C=The specific heat capacity of water is 4.18 J/g/°C
ΔT= 80-35=45C
First you need to convert liters to grams 1 liter = .001 gram so 20KL =20000000 grams
Now just plug in the values to solve for Q
20Mg*4.18 J/g/°C*45C= 3762000000 j = 3762000kj