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## How would you summarize the second Law of Thermodynamics?

Truong-Son N.
Featured 4 months ago

Levine puts it rather elegantly:

As some wit has put it: "The first law says you can't win; and the second law says you can't break even." (Physical Chemistry, Levine, pg. 79)

Or, in my own words,

The first law says energy is conserved, and the second law says the change in entropy of the universe is nonnegative.

I like Levine's wording better, honestly.

• The first law of thermodynamics states:

Energy cannot be created nor destroyed. That is, energy is conserved in the universe.

Therefore, you "can't win" - you can't obtain more energy than you put in. It all has to be accounted for (since the universe is so big, it is quasi-closed).

• The Kelvin-Planck statement of the second law of thermodynamics is (Physical Chemistry, Levine, pg. 79):

It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings.

In other words, the work output can be no greater than the heat input, i.e. the energy efficiency $e$ can never be #100%#:

$e = | w \frac{|}{{q}_{H}}$, #" "" "0% < e < 100%#

where $w$ is the work output from the engine, and ${q}_{H}$ is the heat input from the hot reservoir into the engine.

Another way of saying it is therefore, "you can't break even!"

Put in yet another way - since some heat input is always lost outwards (towards the cold reservoir in the engine scheme), there is always more heat to give more motion.

Entropy, the amount of energy dispersal, increases with more motion present, so the change in entropy of the universe must be nonnegative.

## Two smaller spherical raindrops collide and merge into one larger spherical raindrop. What is the radius and surface energy of this larger drop, compared to the two smaller ones?

MetaPhysik
Featured 3 months ago

Radius increases by 26% while the surface energy drops 21%

#### Explanation:

For a sphere, its volume and surface A area are:
#V=(4pi)/2R^3; A = 4pir^2#
${E}_{s}$ = surface energy = gamma

A) Volume consideration.

Let
${r}_{1} =$radius of small droplet 1
${r}_{2} =$radius of small droplet 2
$R =$radius of merged droplet

Because density of water and total mass of the droplets do not change, the volume before and after the merger must be the same.

Volume of two small droplets = Volume of merged droplet

$\frac{4 \pi}{3} {r}_{1}^{3} + \frac{4 \pi}{3} {r}_{2}^{3} = \frac{4 \pi}{3} {R}^{3}$

${R}^{3} = {r}_{1}^{3} + {r}_{2}^{3}$
#R = (r_1^3+r_2^3)^(⅓)#

If the small droplets have the same size, then ${r}_{1} = {r}_{2} = r$,

#R= (2r^3)^(⅓) = root(3)2 r = 1.26r#

$\frac{\Delta r}{r} =$(1.26r-r)/r = 0.26 = 26%

The radius of the merged droplet is 26% larger.

(B) Surface area consideration , assuming small droplets have identical radius.

Let
$\sigma =$ surface tension
${E}_{s} =$ surface energy $= \sigma A = 4 \pi {r}^{2} \sigma$
${E}_{1} = {E}_{2} = 4 \pi {r}^{2} \sigma =$ surface energy of small droplets
${E}_{3} = 4 \pi {R}^{2} \sigma$ = surface energy of the merger droplet

The percentage change in surface energy is:

$\frac{\Delta E}{{E}_{1} + {E}_{2}} = \frac{{E}_{3} - \left({E}_{1} + {E}_{2}\right)}{{E}_{1} + {E}_{2}} = \frac{{E}_{3} - 2 {E}_{1}}{2 {E}_{1}} = {E}_{3} / \left(2 {E}_{1}\right) - 1$

$\because R = \sqrt[3]{2} r$
#therefore E_3/(2E_1)-1=(4piR^2sigma)/(8pir^2sigma)-1= (4pisigma(root(3)2r)^2)/(8pir^2sigma)-1=-0.21 =- 21% #

The surface energy has reduced by 21%.

## What are Gs? And why do they increase with speed?

Mark C.
Featured 3 months ago

Firstly, the previous answer (Steve’s) is correct, I just wanted to add something on the end in relation to speed.

#### Explanation:

For this to make sense you (probably) need to alter your understanding of how forces work a little.

When an object turns a corner, or moves in circular motion there only needs to be one force acting (there may be more, but only 1 is required) and that force is centripetal i.e. acting towards the centre of the circle. A good example is the orbit of the moon (for this argument, a circle) around the earth. The only force acting is gravity and that acts towards the earth (centre of mass, close to the earth, but again we’ll ignore that for now) causing it to accelerate.

Now, to develop this idea we need to clarify what we mean by an acceleration, and Newton’s 1st law. An object can have a speed in a straight line, but this tells you nothing about the direction of travel. For that, you need a concept called velocity which describes both the speed and direction (in Physics we call these vectors.)

Acceleration is defined as a change in velocity per unit time, and that allows either the speed or the direction to alter as time goes by. In other words, turning a corner at constant speed means you are accelerating (because the direction changes) and thus requires a net force. Newton’s 1st law states that an object will continue in uniform motion (a straight line at the same speed) unless acted upon by an external force. So if there is no net force, an object keeps going in the same direction at the same speed.

Secondly, why does this not tally with your experience of being in a car on a roundabout? You seem to feel a force outwards, that people call centrifugal, but is in fact just the effect of mass not wanting to change where it is (or where it is going) called inertia. Masses, in effect, resist acceleration, including turning their direction, and the greater the mass, the more resistance there is.

Now we can finally relate the question (say an F1 car taking a high speed turn) to the so called “g force” experienced by the driver. As the mass requires a force to turn it, given by $F = \frac{m {v}^{2}}{r}$. This shows that as speed increases the force required to turn the mass increases exponentially (as a squared term) and as the radius reduces (the turn on the track becomes ‘tighter’) so the force also rises. This required force (on a level track) comes from the friction between tyre and track and points inwards (centripetally) towards the inside of the turn.

The inertia of the mass (driver) creates an apparent force pointing outwards, but we now know this is just mass being “stubborn” and not willing to accelerate. This “force” is described in units of “g” or “G” to compare it to the force we experience as gravity, in other words in units of 9.81 Newtons for every kilogram of mass.

Sorry for the long answer, but it isn’t simple! Any clearer?

## State the factors that influence gravity within the surface of the earth ?

1s2s2p
Featured 2 months ago

Your altitude and the position of the centre of gravity of the Earth.

#### Explanation:

The equation for $g$ on Earth is given by:
${g}_{E} = \frac{G {M}_{E}}{r} ^ 2$, where:

• ${g}_{E}$ = acceleration due to free fall on Earth ($m {s}^{-} 2$)
• $G$ = gravitational constant (#~6.67*10^-11Nm^2kg^-2#)
• ${M}_{E}$ = mass of the object (#~5.972*10^24kg#)
• $r$ = distance between the centre of gravities of the two objects ($m$)

Since $G$ and ${M}_{E}$ are constants $g \propto \frac{1}{r} ^ 2$

$r$ is possible to change even without you moving since many things like magma flow through the Earth which have very tiny changes in the position of the centre of gravity that will slightly change $r$.

Let's say you were 7000km away from the centre of gravity from the Earth:
$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{7 \cdot {10}^{6}} ^ 2 = 8.129 m {s}^{-} 2$

Now 5000km:
$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{5 \cdot {10}^{6}} ^ 2 = 15.93 m {s}^{-} 2$

$r$ is usually 6371km

$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{6371 \cdot {10}^{3}} ^ 2 = 9.813646787 m {s}^{-} 2$

But if $r$ was to get 1m less due to movements in the Earth's mantle (for example):
$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{\left(6371 \cdot {10}^{3}\right) + 1} ^ 2 = 9.813643707 m {s}^{-} 2$

A 1m change has a slightly small change in the value for $g$

Also, $r$ could change if the ground was to become raised or lowered,movements of liquids like magma can raise and lower the ground, changing the distance between the person and the centre of gravity for the Earth (assuming that hasn't changed). If the ground was to sink by $1 m$ then $g$ will become:
$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{\left(6371 \cdot {10}^{3}\right) - 1} ^ 2 = 9.813649868 m {s}^{-} 2$ as you can see, $g$ has increased by a very small amount, but not enough to have any noticeable effect on us.

## What are the evidences in favor of the Nuclear Shell Model ?

Aritra G.
Featured 2 months ago

There are multiple experimentally determined facts that indicate the shell-like structure of the atomic nucleus. Some of them are listed below :

1) Nuclei with Z and N (or both) equal to 2, 8, 28, 20, 50, 82 and 126 are extra stable. These numbers are called the magic numbers and this stability variation is analogous to stability of 2, 8, 18, 32 etc electrons in electronic shells.

2) Nuclei with N = a magic number have smaller neutron absorption cross section as compared to their immediate neighbours indicating a more stable nucleus for those values of N.

3) The electric quadrapole moment is zero for a nucleus with N or Z equal to a magic number indicating the spherical symmetry analogous to that in closed shells of atoms.

4) Nuclear species with one neutron excess of a magic number readily emit a single neutron indicating that the extra neutron is loosely bound to the rest of the nucleus.

5) Nuclei with magic number of nucleons of a particular type have more number of stable isotopes and isotones.

## Question #8801d

A08
Featured 2 months ago

This is what I get

#### Explanation:

Voltage vs time relationship of a charging $R C$ circuit is shown in the figure above and is given as

$V \left(t\right) = {E}_{s} \left(1 - {e}^{- \frac{t}{\tau}}\right)$ ......(1)
where ${E}_{s}$ is the DC supply across $R C$ circuit of time constant $\tau$.

It is given that ${E}_{s} = {V}_{C} - {0}_{\text{ref}} = 8.262 - 2.385 = 5.877 \setminus V$

We also know that charge held in a capacitor can be written as

$Q = C V$.

Therefore (1) becomes

$Q \left(t\right) = C {E}_{s} \left(1 - {e}^{- \frac{t}{\tau}}\right)$ ......(2)

Total charge would be maximum at $t = \infty$. From (2) we get

${Q}_{\max} = C {E}_{s}$ ........(3)

Given condition is Charge on the capacitor must be $= \frac{15}{16} {Q}_{\max}$ $\implies$ at what time $t$ or how many times $R C$.
Let the desired time be $= x \tau$

$\frac{15}{16} {Q}_{\max} = C {E}_{s} \left(1 - {e}^{- \frac{x \tau}{\tau}}\right)$ .......(4)

From (3) and (4) we get

$\frac{15}{16} = \left(1 - {e}^{- x}\right)$

Inserting various values in (1) we get

$V \left(t\right) = 5.877 \times \frac{15}{16} = 5.510 \setminus V$

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