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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
answer was not correct it was 2440kJ according to solution manual.
any way looks to be correct and close to the answer
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