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Reika [66]
3 years ago
15

Some element can be either solid or liquid. At the melting point, the liquid has 8 × 10-22 J more enthalpy per atom than the sol

id. The liquid also has 6 × 10-24 J/K more entropy per atom than the solid.
1) What is the latent heat per mole of the phase transition between the solid and the liquid?

? J/mole

2) At what temperature does the melting transition occur?

Tmelting = ? K
Physics
1 answer:
OlgaM077 [116]3 years ago
3 0

Answer:

481.76 J/mol

133.33 K

Explanation:

N_A = Avogadro's number = 6.022\times 10^{23}

Change in enthalpy is given by

\Delta H=8\times 10^{-22}\times 6.022\times 10^{23}\\\Rightarrow \Delta H=481.76\ J/mol

Entropy is given by

\Delta S=6\times 10^{-24}\times 6.022\times 10^{23}\\\Rightarrow \Delta S=3.6132\ J/mol K

Latent heat of fusion is given by

L_f=\Delta H\\\Rightarrow L_f=481.76\ J/mol

The latent heat of fusion is 481.76 J/mol

Melting point is given by

T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K

Melting occurs at 133.33 K

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ladessa [460]

Answer:

A measure of the ability of a material to transfer heat.

Explanation:

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4 0
3 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
3 years ago
If it takes 50n of force to lift a 450n what is the ma of the machine
otez555 [7]

If the machine is 100% efficient, then its
Mechanical Advantage is (450/50) = 9 .

If the machine is less than 100% efficient,
then the MA is more than 9 .

7 0
3 years ago
A young parent is dragging a 65 kg (640 N) sled (this includes the mass of two kids) across some snow on flat ground, by means o
Delicious77 [7]

Answer:

b) N = 560 N, c)  fr = 138.56 N, d)  μ = 0.247

Explanation:

a) In the attachment we can see the free body diagram of the system

b) Let's write Newton's second law on the y-axis

              N + T_y -W = 0

              N = W -T_y

let's use trigonometry for tension

             sin θ = T_y / T

             cos θ = Tₓ / T

             T_y = T sin θ

             Tₓ = T cos θ

we substitute

              N = W - T sin 30

we calculate

              N = 640 - 160 sin 30

              N = 560 N

c) as the system goes at constant speed the acceleration is zero

X axis

              Tₓ - fr = 0

               Tₓ = fr

we substitute and calculate

              fr = 160 cos 30

              fr = 138.56 N

d) the friction force has the formula

             fr = μ N

             μ = fr / N

we calculate

             μ = 138.56 / 560

             μ = 0.247

4 0
3 years ago
In 2014, about how far in meters would you have to travel on the surface of the Earth from the North Magnetic Pole to the Geogra
pogonyaev

Answer:

267.07 km

Explanation:

We have given the radius of the earth = 6378.1 km

In 2014 the difference between the magnetic north pole and geographical north pole is 2.40°

2.40°=2.40\times \frac{\pi }{180}=0.041866 radian

We know that linear distance is given by S=R\Theta =6378.1\times 0.041866=267.07km

So we have to travel 267.07 km in going from magnetic north pole to geographic north pole

3 0
3 years ago
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