find the energy that must be transferrred as heat to evaporate 1 kg of water at 25 degree celsius.answer is 2440 kJ.how come?please solve this

ΔQ=msΔT;      m=1kg; ΔT=100-25=75°C; s=specific heat for water=4184 J/kg/°C

So, ΔQ=1* 4184 *75=313800J              -----  Answer

ACTUALLY THE ANSWER THAT YOU HAVE GIVEN IS WRONG 

   IF THE WATER WANT TO FULLY EVAPORATE FIRST IT WILL MAKE ITS TEMPRATURE TO 100 DEGREE CELICIUS THEN IT WILL CHANGE ITS STATE TO GASEOUS FORM BY PHASE TRANSFORMATION SO TOTAL HEAT REQUIRED

    ms(T₂-T₁) +  mL 

   where ''m'' is the mass and ''s'' is heat capacity of water ''L'' is latent heat of vapourisation ''T₁'' is intial temprature and ''T₂'' is final temprature

  S = 4200J/KG/DEGREE CELCIUS

  L  = 2.26 X 10⁶ J/KG/DEGREE CELCIUS 

  T₁ = 25 DEGREE CELCIUS

  T₂  = 100 DEGREE CELCIUS

SUBSTITUTE THE VALUES IN THE EQUATION AND GET YOUR ANSWER  WHICH COMES OUT TO BE 2575KJ

  IF YOU ARE SATISFIED PLZZ APPROVE MY ANSWER

Approved

answer was not correct it was 2440kJ according to solution manual.

any way looks to be correct and close to the answer