Answer:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Explanation:
If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (
), in joules per gram-Kelvin, by the following model:
![s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }](https://tex.z-dn.net/?f=s_%7B2%7D%20-%20s_%7B1%7D%20%3D%20m%5Ccdot%20c_%7Bw%7D%20%5Ccdot%20%5Cint%5Climits%5E%7BT_%7B2%7D%7D_%7BT_%7B1%7D%7D%20%7B%5Cfrac%7BdT%7D%7BT%7D%20%7D)
(1)
Where:
- Mass, in kilograms.
- Specific heat of water, in joules per kilogram-Kelvin.
,
- Initial and final temperatures of water, in Kelvin.
If we know that
,
,
and
, then the change in entropy for the entire process is:
![s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}](https://tex.z-dn.net/?f=s_%7B2%7D%20-%20s_%7B1%7D%20%3D%20%281%5C%2Ckg%29%20%5Ccdot%20%5Cleft%284190%5C%2C%5Cfrac%7BJ%7D%7Bkg%5Ccdot%20K%7D%20%5Cright%29%5Ccdot%20%5Cln%20%5Cfrac%7B288.15%5C%2CK%7D%7B373.15%5C%2CK%7D)
![s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}](https://tex.z-dn.net/?f=s_%7B2%7D%20-%20s_%7B1%7D%20%3D%20-1083.112%5C%2C%5Cfrac%7BJ%7D%7Bkg%5Ccdot%20K%7D)
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Answer:
D & E
Explanation:
I think this is dealing with latent heat and D & E would be the range where you will find solid and liquid phases in equilibrium, cuz it starts as gas at from A to B, B to C is gas and liquid equilibrium, C to D is liquid, D to E solid and liquid, and then E to F is solid.
The statement that correctly compares protons, electrons, and neutrons is, "Quarks are present in protons and neutrons but not in electrons". Quarks are tiny particles which compose the neutrons and protons. These quarks in protons and neutrons are linked together by gluons.
Remains the same because temperature can't change energy