Calculate the change in internal energy of 2kg of water at 90 degree celcius when it is changed to 3:30m3 of steam at 100 degrees. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is 2:26 x 10^6 J/kg.

Question

Calculate the change in internal energy of 2kg of water at 90 degree celcius when it is changed to 3:30m3 of steam at 100 degrees. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is 2:26 x 10^6  J/kg.

anonymous · Answer

maybe 4604000 J 

just did it using some of my own values :p

mos1635 · Answer

please elaborate...

anonymous · Answer

m*c*t=Q

2*4200*10=84000

m*l=Q

2*2.26x10^6=4520000

Total= 4520000+ 84000

4604000J

anonymous · Answer

argh i dont know if has to do something with the Charles and boyles laws

mos1635 · Answer

that is exactly my thinking but that 3:30m3 is confusing.....

mos1635 · Answer

f 2Kgr of Water terned in gas in 
P=1 atm 
T=100+273
R=0.082 
n=1000/18 
should have volule V=1.699 m3

anonymous · Answer

n=2000/18

mos1635 · Answer

oooooppsssss

mos1635 · Answer

and i did say 2 Kgr of water!!!! LOL

mos1635 · Answer

4604000J it is then...

anonymous · Answer

pv=nRT

is ideal gas equation

i doubt if water ( steam) is an ideal gas ;)

mos1635 · Answer

in higt temp and low densidy that is an acceptable aproximation

mos1635 · Answer

aply PV=nRT gives 3.39844444 m3 close enough

anonymous · Answer

@mos1635  @hashsam1  hello guys why are u relating the solution to PV = nRt and how do u arrive at it values

mos1635 · Answer

i was just cheking if given value of volume had any significance to the question. No need to conclude that in your answer.

mos1635 · Answer

if we ignore given volume we have
1st
water from 90 to water 100
Q1=m*c*ΔΤ1
m=2 Kgr
c=4200 joule/Kgr*K
ΔT1=10
2nd
water from 100 to steam 100
Q2=m*l
m= 2Kgr
l=2.26 x 10^6 J/kg.

anonymous · Answer

thanks @mos1635