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4 years ago

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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?

1 answer:

Sunny_sXe [5.5K]4 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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One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f

Andreas93 [3]

Answer:

Part a)

$\theta_2 = 15 degree$

Part b)

$\Delta t = 2.88 s$

Explanation:

Part a)

In order to have same range for same initial speed we can say

$R_1 = R_2$

$\frac{v^2 sin2\theta_1}{g} = \frac{v^2 sin2\theta_2}{g}$

so after comparing above we will have

$\theta_1 = 90 - \theta$

so we have

$75 = 90 - \theta_2$

$\theta_2 = 15 degree$

Part b)

Time of flight for the first ball is given as

$T_1 = \frac{2vsin\theta}{g}$

$T_1 = \frac{2(20)sin75}{9.81}$

$T_1 = 3.94 s$

Now for other angle of projection time is given as

$T_2 = \frac{2(20)sin15}{9.81}$

$T_2 = 1.05 s$

So here the time lag between two is given as

$\Delta t = T_1 - T_2$

$\Delta t = 3.94 - 1.05$

$\Delta t = 2.88 s$

5 0

3 years ago

A mobile phone is 35% efficient. Over half an hour 11 kJ of energy is transferred to the phone.

kramer

<h3><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u></h3>

Energy Transferre=11KJ
Efficiency=35%

<h3>☆Usefully transferred energy:-</h3>

$\\ \sf\longmapsto 35\%\:of 11$

$\\ \sf\longmapsto 35\%\times 11$

$\\ \sf\longmapsto \dfrac{35}{100}\times 11$

$\\ \sf\longmapsto \dfrac{385}{100}$

$\\ \sf\longmapsto 3.85KJ$

$\\ \sf\longmapsto 3850J$

6 0

3 years ago

The electrons that produce the picture in a

vredina [299]

Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled

Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²

Answer: 1.95 x 10¹⁴ m/s²

4 0

3 years ago

As soon as a traffic light turns green, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.

Ganezh [65]

Answer: (a) The bicycle is ahead of the car for 4 s.

(b) The bicycle leads the car by the maximum distance of 55 m.

Explanation:

(a)

Use the equation of the motion to calculate the time taken by the car.

$v=u+at$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .

Put u=0, v=22.3 m/sec and a=4.02 m/sec^2.

$22.3=0+4.02t$

$t=\frac{22.3}{4.02}$

t= 5.5 s

Use the equation of the motion to calculate the time taken by the bicycle.

$v=u+at_{1}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.

$8.94=0+5.81t_{1}$

$t_{1}=\frac{8.94}{5.81}[tex]t_{1}=1.5 s$

Calculate the time interval for which the bicycle is ahead of the car.

$t-t_{1}= 5.5 s - 1.5s$

$t-t_{1}= 4s$

Therefore, the bicycle is ahead of the car for 4 s.

(b)

Use the equation motion to calculate the distance covered by the car.

$S=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .

Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.

$S=(0)t+\frac{1}{2}(4.02)(5.5)^{2}$

$S= 60.8 m$

Use the equation motion to calculate the distance covered by the bicycle.

$S_{1}=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.

$S_{1}=(0)t+\frac{1}{2}(5.18)(1.5)^{2}$

$S_{1}= 5.8 m$

Calculate the maximum distance covered by the bicycle to lead the car.

$S-S_{1}=60.8-5.8=55m$

Therefore, the bicycle leads the car by the maximum distance of 55 m.

6 0

3 years ago

In a local park, a pine cone falls off a tree branch (starting at rest) and falls to the ground. We will model the pine cone hit

Liono4ka [1.6K]

Answer:

1)) ΔU = -8.96 J, 2) k = 8.18 10⁴ N / m, 3) v = 8.47 m / s

Explanation:

For this exercise we will use conservation of energy.

Starting point. Point where the pineapple comes out

Em₀ = U = m g h

where the reference frame is placed on the ground

Final point. Point where pineapple stops

Em_f = K_e + U = ½ k y² + m g y

1) the change in gravitational potential energy is

ΔU = U_f - U₀

ΔU = m g y - m g h

ΔU = mg (y-h)

let's calculate

ΔU = 0.116 9.8 (0.0148 - 7.9)

ΔU = -8.96 J

The negative sign indicates that the energy decreases

2) let's use energy conservation

Em₀ = Em_f

mg h = ½ k y² + mg y

k = mg (h-y) $\frac{2}{y^2}$

let's calculate

k = 0.116 9.8 (7.9 - 0.0148) $\frac{2}{0.0148^2}$

k = 8.18 10⁴ N / m

3) we use the same starting point and as the end point we use this height (y₂ = 4.24 m)

Em_{f2} = K + U = ½ m v² + mg y₂

energy is conserved

Em₀ = Em_{f2}

mgh = ½ m v² + m g y₂

v = $\sqrt{ 2g(h-y_2)}$

let's calculate

v = $\sqrt{ 2 \ 9.8 \ (7.9-4.24)}$

v = 8.47 m / s

6 0

3 years ago