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Possible conversion and units problems 
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#1
May312, 12:54 AM

P: 15

I have figured out most i needed from my earlier entropy questions but now i'm faced with a different problem...
dS=δQ/T that's the formula for the equation. I just did 60 grams of water at 273 K and a room at 298 K so the system is 273 K and the surroundings are 298K Using the forumula for thermal energy (m*Cp*ΔT) i found that the thermal energy is .6285 (what are the units I measure this in?) by doing .06 kg*4.19 J/g degrees C(Cp of water)*25 K(change in temperature i.e. 298273) then i just plugged it in so dS=.6285/273, so dS = .0023021 J/K (or is it J/ Kg K?) and the same for the surrounding but in the negative. dS = .6825/298, so dS = .0022902 J/K? so although this shows that entropy overall increased because of the ratio of increase from the system to decrease from the surroundings, I'm concerned because this isn't as infinitesimaly small of a change that it should be I suppose. So I'm wondering where I went wrong, units, conversions, whatever. like the specific heat capacity in J/g degrees C. Should this have been converted to J/kg K? If so, how? Please do correct me. Thanks a ton! 


#2
May412, 07:49 AM

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This is only going to be approximately right because the temperature changes continuously. To get a more accurate answer you need to do the integration.
.06 kg*4.19 J/g degrees K*25 K = 6.285 kg J / g = 6285 J So it looks to me that you are underestimating the increase by a factor of 10,000. What makes you think that the answer is too big? I.e. why do you think you should have any feel for what an infinitesimal change in entropy would look like in numbers? 


#3
May512, 10:41 AM

P: 15

I see it now, before I was using my units wrong. I didn't convert from kg to g, which I see why I should. I'm still a little confused because the Cp of water is 4.19 j/g degrees C, not K. So should i convert that as well, and how if I do? Obviously the thing that's tripping me up is units. And i'm confused because I was thinking for an infinitesimal change, it would have to be some incredibly small decimal. Is that not the case?



#4
May512, 03:35 PM

P: 168

Possible conversion and units problems
c_{p} = 4.19 J/(gC). Note J/g, not J/(KgC). but you have used mass m = .06 Kg. c_{p} is in gm but mass is in Kg. I'll use m = 60 gm to compute δQ, which would be in J. ΔT doesn't seem to matter, because 273 added to both low and high temp. But IIRC, our texts use T=285.5^{o}K in the entropy equation, average of the two temps, because T on the bottom should be at a constant temperature. I get dS = 22.01 J/K. [I'm assuming your conversion number 4.19 is correct] 


#5
May512, 08:12 PM

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You can't judge whether an answer is large or small just from its numerical value  you have to consider the units, and you also need some context to appreciate what it means. I.e., large or small compared with what? 


#6
Jun1612, 06:50 AM

P: 2

unit and conversions of engineering quantities visit
http://marinesite.blogspot.in/2012/0...nversions.html 


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