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3 years ago

How does the entropy of steam compare to the entropy of ice?

Physics

2 answers:

baherus [9]3 years ago

6 0

Answer:

One of the thermodynamic principles says that the entropy increases when the temperature increases, and we have the relation:

dS/dT > 0.

Now, we know that the steam will have more temperature than the ice, so we can say that the entropy of the steam is bigger than the one of the ice.

You also can see in a more "simplistic" concept, entropy is viewed as the amount of "disorder". Now, then the ice has a defined shape where each particle kinda maintains their position, the case is different for gas (the steam in this case) where the particles are almost free, then in the gas, the disorder is bigger than in the ice, and the entropy is also bigger.

Dmitriy789 [7]3 years ago

3 0

The answer is; The entropy of steam is larger than because it is more disordered than ice

Entropy is synonymous to the degree of disorder or randomness of molecules in a system. The molecules of steam are far apart from each other and move randomly in the system colliding with each other. Those of ice has less kinetic energy, vibrate more or less in a fixed position in the structiure, and are arranged in a orderd fashion

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A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

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<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

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Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

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Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

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