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sashaice [31]
3 years ago
9

A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C. What is the entropy change of the sample if the latent hea

t of vaporization of water is 2.26 x 10⁶ J/kg?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

The entropy change of the sample of water =  6.059 x 10³ J/K.mol

Explanation:

Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol

Mathematically, entropy is expressed as

ΔS = ΔH/T....................... Equation 1

Where ΔH = heat absorbed or evolved, T = absolute temperature.

<em>Given:  If 1 mole of water = 0.0018 kg,</em>

<em>ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.</em>

<em>T = 100 °C = (100+273)  K = 373 K.</em>

<em>Substituting these values into equation 1,</em>

<em>ΔS =2.26x 10⁶/373</em>

ΔS = 6.059 x 10³ J/K.mol

Therefore the entropy change of the sample of water =  6.059 x 10³ J/K.mol

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The band of stability curves upward at high atomic numbers due to the fact that excess of neutrons are required due to the repulsion between protons.

Atomic number is the number of protons. As the number of protons (atomic number) increase, the electrical repulsion force, due to the same sign of the protons inside the nucleus, becomes more dominant compared to the nuclear force attractions, then the nucleus needs more neutrons to gain stability.The presence of more neutrons decrease the density of protons which decreases the repulsion among them.


3 0
3 years ago
A spring of force constant 285.0 N/m and unstretched length 0.230 m is stretched by two forces, pulling in opposite directions a
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Answer: W = 0.3853 J, e = 0.052 m

Explanation: Given that,

K =285.0N/M , L = 0.230m , F = 15N , e = ?

F = Ke

15 = 285 × e

e = 15÷ 285

e =0.052 m

e + L = 0.052 + 0.230

= 0.282m ( spring new length )

Work needed to stretch the spring

W = 1/2ke2

W = 1/2 × 285 x 0.052 × 0.052

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8 0
3 years ago
2. An object with moment of inertia ????1 = 9.7 x 10−4 kg ∙ m2 rotates at a speed of 3.0 rev/s. A 20 g mass with moment of inert
svetoff [14.1K]

Answer:

2.85 rad/s

Explanation:

5 cm = 0.05 m

20 g = 0.02 kg

When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:

I_2 = 1.32\times10^{−6} + 0.02 * 0.05^2 = 0.513\times10^{-4} kgm^2

So the total moment of inertia of the system of 2 objects after the drop is:

I = I_1 + I_2 = 9.7\times10^{-4} + 0.513\times10^{-4} = 0.0010213 kgm^2

From here we can apply the law of angular momentum conservation to calculate the post angular speed

\omega_1 I_1 = \omega_2 I

\omega_2 = \omega_1 \frac{I_1}{I} = 3 \frac{9.7\times10^{-4}}{0.0010213} = 2.85 rad/s

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3 years ago
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6 0
2 years ago
Which particles are located in the nucleus of the atom? (2 points)
pentagon [3]

Answer:

Neutrons and protons are located in the nucleus of the atom.

Explanation:

And electrons are in the electron cloud.

8 0
3 years ago
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