Possible conversion and units problems

  • Thread starter infinitylord
  • Start date
  • Tags
    Units
In summary, the individual is discussing their calculations for thermal energy and entropy, using the formula δQ/T and the specific heat capacity of water. They realize they made a mistake with their units and are concerned about the size of their calculated entropy change, but they are unsure of what the correct units and conversions should be. They also question the concept of an infinitesimal change and its numerical value.
  • #1
infinitylord
34
1
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. 298-273)
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!
 
Physics news on Phys.org
  • #2
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
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
infinitylord said:
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. 298-273)
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!

I see one problem in the numbers you have used,
cp = 4.19 J/(gC). Note J/g, not J/(KgC). but you have used mass m = .06 Kg.
cp 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=285.5oK 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]
 
Last edited:
  • #5
infinitylord said:
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?
A 1C change is also a 1K change, so no need for a conversion there.
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?
 

1. How do I convert units of length?

To convert units of length, you first need to know the conversion factor between the two units. Then, multiply the original measurement by the conversion factor to get the converted measurement. For example, to convert 10 inches to centimeters, you would multiply 10 by 2.54 (the conversion factor between inches and centimeters) to get 25.4 centimeters.

2. How can I convert units of temperature?

To convert units of temperature, you will need to use either the Fahrenheit, Celsius, or Kelvin scale. To convert from Fahrenheit to Celsius, subtract 32 from the Fahrenheit temperature and then multiply by 5/9. To convert from Celsius to Fahrenheit, multiply the Celsius temperature by 9/5 and then add 32. To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature.

3. How do I convert units of volume?

To convert units of volume, you first need to know the conversion factor between the two units. Then, multiply the original measurement by the conversion factor to get the converted measurement. For example, to convert 1 liter to milliliters, you would multiply 1 by 1000 (the conversion factor between liters and milliliters) to get 1000 milliliters.

4. How can I convert units of mass?

To convert units of mass, you first need to know the conversion factor between the two units. Then, multiply the original measurement by the conversion factor to get the converted measurement. For example, to convert 5 kilograms to grams, you would multiply 5 by 1000 (the conversion factor between kilograms and grams) to get 5000 grams.

5. How do I convert between metric and imperial units?

To convert between metric and imperial units, you will need to use conversion factors and formulas specific to the type of unit being converted. It is important to double check the accuracy of the conversion factor and to make sure you are using the correct formula for the specific type of unit being converted. It may also be helpful to use a conversion calculator or reference chart to ensure accuracy.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
931
  • Introductory Physics Homework Help
Replies
4
Views
763
  • Introductory Physics Homework Help
Replies
12
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
999
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
956
  • Introductory Physics Homework Help
Replies
2
Views
707
  • Engineering and Comp Sci Homework Help
Replies
6
Views
737
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
Back
Top