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OpenStudy (anonymous):

Calculate the change in internal energy of 2kg of water at 90 degree celcius when it is changed to 330m^3 of steam at 100oC. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is 226*10^6 J/kg

OpenStudy (anonymous):

Just Remember two things \[\large Q = Q \space ( H_2O _{(l->l)(90^o \to 100^o C)})+ Q \space(H_2O _{(l->g)(100^o \to 100^o C)})\] \[\large Q \space ( H_2O _{(l->l)(90^o \to 100^o C)})= m*c_{water}*\Delta T\] \[\large Q \space ( H_2O _{(l->g)(100^o \to 100^o C)})= m*L\] m-->mass L-->Latent heat ΔT -->Change in temperature Also to find mass of steam mass = (density)(volume) density of steam = 1000 kg/m^3 Now try :)

OpenStudy (anonymous):

l-->liquid g-->gas

OpenStudy (anonymous):

@Prof.Grace , Any problem/doubt??

OpenStudy (anonymous):

shouldn't the mass of water and steam be equal

OpenStudy (anonymous):

I forgot to mention that c-->specific heat

OpenStudy (anonymous):

c_water = 4.186 joule/gram °C

OpenStudy (anonymous):

Be careful with units. :)

OpenStudy (anonymous):

@Prof.Grace , There is something wrong with the question. The volume of steam is too high. Check the question again

OpenStudy (anonymous):

that's exactly what i was wondering...but that's the question

OpenStudy (anonymous):

I found a similar question on net. See here -> http://www.assignmentexpert.com/free-questions/question-on-physics-molecular-physics-and-thermodynamics-8650.html

OpenStudy (anonymous):

Here Volume is 3.30 m^3

OpenStudy (anonymous):

okay, i can see another version of the question - it says 3.3m^3

OpenStudy (anonymous):

Q = (2*4.186*10) + (303.03*2.26*10^6)

OpenStudy (anonymous):

= 83.72 + 684.85*10^6 = 6.5 * 10^8 please help me cross check it - it seems wrong to me

OpenStudy (anonymous):

Q = (2*4.186*10)*10^3

OpenStudy (anonymous):

684.85*10^6 -->100% wrong

OpenStudy (anonymous):

wow that was very frank

OpenStudy (anonymous):

Found the mistake 2.26 X 10^6 J/kg -->This is the right latent heat of ice Your question has lot of decimal mistake Also I think the mass should be 2 kg as you said

OpenStudy (anonymous):

I got it. The water is partially converted to steam. I got it now.

OpenStudy (anonymous):

Also density of steam is 0.6 kg/m3

OpenStudy (anonymous):

Now find mass of steam from given volume

OpenStudy (anonymous):

the answer should be 4.6 * 10^6

OpenStudy (anonymous):

or 4.6 mJ

OpenStudy (anonymous):

Exactly :D

OpenStudy (anonymous):

nice job...***smiles***

OpenStudy (anonymous):

Suggestion -->Your question had lot of decimal related errors which really made us troll this long BTW, did you get the answer or you are telling me the final answer?

OpenStudy (anonymous):

that was what i got

OpenStudy (anonymous):

Very good. Keep these standard values in mind like density of water, steam and specific heat of water always. Many a time they won't be given in the question. Good luck :D

OpenStudy (anonymous):

kk

OpenStudy (mos1635):

note: mass of liquid is always=to mass of steam (that is if all liquid is evaporeted) in same cases if not enough heat is given, onlt a fraction of liquid evaporate and then M(g)<M(l).

OpenStudy (anonymous):

@mos1635 ,Thanks for clarifying. :)

OpenStudy (mos1635):

always nice to see your work :)

OpenStudy (anonymous):

:D .That's too sweet for my ears :D

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