4. An electric iron has a
power rating of 750W
a. How many joules of
electric energy does it
change into heat energy
every second?
b. How many joules of
work can it do in 3
seconds
c. How long does it take
the iron to do 1500J of
work?
5. Use the kinetic particle
theory to explain why a
solid has a definite shape
and liquid has none.​

Answer:

Explanation:

We shall apply Stefan's formula

E = AσT⁴

When T = 300

I₁ = Aσ x 300⁴

When T = 400K

I₂ = Aσ x 400⁴

I₂ / I₁ = 400⁴ / 300⁴

= 256 / 81

= 3.16

I₂ = 3.16 I₁ .

Answer:

Explanation:

I catch the ruler 5.7 cm from lower end that means my reaction time is equal to time of fall of ruler as free fall under gravity .

h = 1/2 gt²

t = [tex]\sqrt{\frac{2h}{g} }[/tex]

= [tex]\sqrt{\frac{2\times 5.7}{9.8} }[/tex]

= 1.078 s

= 1.1 s .

Answer:

The mass of the ice added = 16.71 g

Explanation:

The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.

But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.

Heat gained by the ice = Heat lost by the 326 g of water.

Let the mass of ice be m

The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)

Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT

m = unknown mass of ice

C = Specific Heat capacity of ice = 2.108 J/g°C

ΔT = change in temperature = 0 - (-5.5) = 5.5°C

Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J

Heat used by the ice to melt at 0°C = mL

m = unknown mass of ice

L = Latent Heat of fusion of ice to water = 334 J/g

Heat used by the ice to melt at 0°C = m×334 = (334m) J

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT

m = unknown mass of water (which was ice)

C = Specific Heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 15 - 0 = 15°C

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J

Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J

Heat lost by the water in the calorimeter cup = MCΔT

M = mass of water in the calorimeter cup = 326 g

C = specific heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 20 - 15 = 5°C

Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J

Heat gained by the ice = Heat lost by the 326 g of water.

408.384m = 6,823.18

m = (6,823.18/408.384)

m = 16.71 g

Hope this Helps!!!

Answer:

The same momentum will be the best option.

Explanation:

Let's recall that the force will be express in terms of the momentum. We can write the force as the variation of the momentum over time.

[tex]F=\frac{dp}{dt}[/tex]

This is the force needed to stop the base ball or the bowling ball.

if we will choose the same kinetic energy it would imply an increase of momentum, because of the difference of the masses, and therefore an increase of the force. We do not want this.

Now, if we choose the same momentum the kinetic energy will increase, but the force will the same. We want the less force as possible to stop it, and we have the same at least.

Therefore the same momentum would be the best option.

I hope it helps you!

The best choice to catch and hold the bowling ball will be; with the same momentum

We know that formula for impulse is;

Impulse = Force x Time

And we know that change in momentum is equal to impulse. Thus;

Change in momentum = F × t

ΔP = F × t

F = ΔP/t

This formula represents the force required to stop the baseball or the bowling ball.

Now,  momentum is proportional to the square root of kinetic energy.

Now, since momentum is directly proportional to velocity, while kinetic energy is proportional to the square of the velocity, it means that if kinetic energy is quadrupled, then the momentum will become double.

Now, the collision in the question is completely inelastic and as such, all the bowling balls kinetic energy will be in inelastic collision, the kinetic energy is lost.

Formula for the kinetic energy in terms of the momentum here is;

K = p²/2m

Looking at it overall, we can say that the best choice to catch and hold the bowling ball will be with the same momentum since it results in lesser force.

Read more at;https://brainly.com/question/13994440

Answer:

0.838

Explanation:

The ratio v/c of the speed v to the speed c of light in a vacuum is shown below:

Given that

[tex]\triangle t_0 = 24\ seconds[/tex] = time interval for one revolution

[tex]\triangle t = 44\ seconds[/tex] = time interval measured with speed v

based on the given information, the ratio v/c  of the speed v to the speed c of light in a vacuum is

[tex]\triangle t = \frac{\triangle t_0}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

[tex]{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{\triangle t_0}{\triangle t}[/tex]

Now squaring both the sides

[tex]\frac{v^2}{c^2} = 1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}[/tex]

Now remove the squaring root from both the sides and putting the values

[tex]\frac{v}{c} = {\sqrt{1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}[/tex]

[tex]= {\sqrt{1 - \frac{(24)^2}{(44)^2}[/tex]

= 0.838

Answer:1.89 m

Explanation:

Given

Block travels [tex]0.63\ m[/tex] in first second

It is released from rest i.e. initial speed is zero (u=0)

using

[tex]s=ut+\frac{1}{2}at^2[/tex]

where a=acceleration

here acceleration is the component of gravity on incline plane (say [tex]\theta [/tex])

so

[tex]s_1=\frac{1}{2}\times g\sin \theta (1)^2[/tex]

[tex]0.633\times 2=9.8\sin \theta \times 1^2[/tex]

[tex]\sin\theta =0.1291[/tex]

[tex]\theta =7.41^{\circ}[/tex]

So distance traveled in [tex]2\ sec[/tex]

[tex]s=\frac{1}{2}\times g\sin \theta (2)^2[/tex]

[tex]s=0.5\times 9.8\times \sin (7.41)\times 4[/tex]

[tex]s=2.52\ m[/tex]

So distance traveled in [tex]2^{nd}\ sec[/tex] is

[tex]s-s_1=2.52-0.633=1.89\ m[/tex]

Answer:

The Answer is red is the lowest and violet is the highest frequency

Explanation:

I think that means A, because the red isn't in the question. But I'm sure red is the lowest frequency and violet is the highest in the visible light spectrum

The visible spectrum as it appears to the human eye is that A. the lowest frequency appears red, and the highest frequency appears violet.

Humans can only view a portion of the electromagnetic spectrum and this portion is known as visible light.

The colors in this visible light have different frequencies which include:

Notice how red is the lowest frequency and violet is the highest so we can conclusively say that the lowest frequency appears red, and the highest frequency appears violet.

Find out more at https://brainly.com/question/15091042.

Answer:

f = 7.57 Hz

Explanation:

To find the frequency of the damping oscillator, you first use the following formula for the angular frequency:

[tex]\omega=\sqrt{\omega_o-(\frac{b}{2m})^2}=\sqrt{\frac{k}{m}-(\frac{b}{2m})^2}\\\\[/tex]   (1)

k: spring constant = 2.65*10^4 N/m

m:  mass = 11.7 kg

b: damping coefficient = 4.50 Ns/m

You replace the values of k, m and b in the equation (1):

[tex]\omega=\sqrt{\frac{2.65*10^4N/m}{11.7kg}-(\frac{4.50Ns/m}{2(11.7kg)})^2}\\\\\omega=47.59\frac{rad}{s}[/tex]

Finally, you calculate the frequency:

[tex]f=\frac{\omega}{2\pi}=\frac{47.59}{2\pi}Hz=7.57\ Hz[/tex]

hence, the frequency of the oscillator is 7.57 Hz

Answer:

0.795 kg

Explanation:

Assuming the complete question:

A person with compromised pinch strength in their fingers can only exert a normal force of 6.0 N to either side of a pinch-held  object. What is the mass of the heaviest book this person can hold onto vertically before it slips out of his or her fingers? The  coefficient of static friction of the surface between the fingers and the book cover is 0.60

SOLUTION:

The maximum weight of the book will equal the maximal friction force that can be produced:

m g = 2 f [tex]F{normal}[/tex]

Note that there are two sides of the book, so the friction force equals 2 times the friction force on one side (hence the factor 2).

So the maximum mass of the book is

m = 2 f [tex]F_n[/tex]/ g

m = 2[tex]\times[/tex]0.65[tex]\times[/tex] 6.0N / (9.81N/kg)  

m = 0.795 kg

Answer:

Final velocity of NFL line backer is 16.67 m/s.

Explanation:

From the question, we have following data about the NFL line backer:

Initial Speed of line backer = Vi = 0 m/s (Since, he starts from rest)

Distance covered by NFL line backer = s = 100 m

Time taken by the NFL line backer to complete 100 m sprint = t = 12 s

Acceleration of NFL line backer during sprint = a

Final Velocity of NFL line backer = Vf = ?

First we need to find the acceleration of the NFL line backer. For that purpose we will use 2nd equation of motion:

s = (Vi)(t) + (0.5)at²

using values:

100 m = (0 m/s)(12 s) + (0.5)(a)(12 s)²

100 m/72 s² = a

a = 1.39 m/s²

Now, we use 1st equation of motion to find Vf:

Vf = Vi + at

Vf = 0 m/s + (1.39 m/s²)(12 s)

Vf = 16.67 m/s

Answer:

The answer is 5.05 hours.

Explanation:

If the plane has an airspeed of 203 km/h which only applies for air and not the ground speed, we can subtract the speed of the wind since it is a headwind in the directly opposite direction.

So the speed of the plane becomes 203 - 53.5 = 149.5 km/h which will give us the true airspeed of the plane and the ground speed as well.

From here we can calculate the time it will take to reach the city as

756 km / 149.5 km/h = 5.05 hours.

I hope this answer helps.

Answer:

0.4778 m/s

Explanation:

To solve this question, we will make use of law of conservation of momentum.

We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;

V_x = (12 m/s)(cos(35°)) = 9.83 m/s.

Thus, the horizontal component of the rock's momentum is;

(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.

Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.

Thus, to get the person's speed, we know that; momentum = mass x velocity

Mass of person = 72 kg and we have momentum as 34.405 kg·m/s

Thus;

34.405 = 72 x velocity

Velocity = 34.405/72

Velocity = 0.4778 m/s

Answer: acceleration = 3.27m/s^2

Explanation:

Given that the

Mass M = 44kg

Angle Ø = 15 degree

Coefficient of friction ų = 0.7

Force F = 222N

F - Fr = ma ...... (1)

Where Fr = frictional force

Fr = ųN

N = normal reaction = mg

Fr = ųmgsinØ

Fr = 0.7 × 44 × 9.81 × sin 15

Fr = 78.2N

Substitutes Fr, F and M into equation one.

222 - 78.2 = 44a

143.79 = 44a

Make a the subject of formula

a = 143.79/44

Acceleration a = 3.27 m/s^2

Answer:

P = I²r

Explanation:

ε= IR + Ir

where r is the internal resistance

Explanation:

A substance with a short , medium, free path has improved electron flow resistance and a higher electrical resistance . Heat applications impose more molecular chaos on all materials and shorten the track further, increasing resistance of most materials. So, just refresh the material to expand the course. In certain materials, when cooled to the minimum temperature, the conductivity is substantially increased.

Answer:

1) We can estimate the age of the earth

2) we can calculate the speed of anything

3) we can also calculate gravity, e.t.c.

Explanation:

I could give you more just ask

Answer:[tex]60^{\circ}[/tex]

Explanation:

Given

[tex]\mid\Vec{A}\mid=1[/tex]

[tex]\mid\Vec{B}\mid=2[/tex]

And [tex]A\cdot B=1[/tex]

We know [tex]\vec{A}\cdot \vec{B}=\mid\Vec{A}\mid\mid\Vec{B}\mid\cos \theta[/tex]

Where [tex]\theta[/tex] is the angle between them

Substituting the values

[tex]1=1\times 2\cos \theta[/tex]

[tex]\cos \theta =\dfrac{1}{2}[/tex]

[tex]\theta =60^{\circ}[/tex]

Thus the angle between [tex]A[/tex] and [tex]B[/tex] is  [tex]60^{\circ}[/tex]

Answer:

1. v =  22.2 m/s

2. t = 2.25 seconds

3. h = 27.05 m

4. t = 1.16 seconds

Explanation:

The questions involve motion under the influence of gravity

1. Using the formula v² = u² + 2gh

where u = 1.6 m/s; g= 9.81 m/s²; h = 25 m; v = ?

v² = (1.6)² + 2 * 9.81 * 25

√v² = √493.06

v =  22.2 m/s

2. Using h = ut + 1/2 gt²

where h = 25 m; u = 0 (since velocity on reaching the ground is zero); g = 9.81 m/s²; t = ?

therefore, h = 1/2 gt²

making t subject of the formula, t = √ (2*h /g)

t = √ (2 * 25 / 9.81)

t = 2.25 seconds

3. Time of travel for the blue ball, t = 2 - 0.4 = 1.6s

using h = ut - gt²

u = 24 m/s; t = 1.6 s; g = 9.81 m/s²

note: since the ball is travelling against gravity, g is negative

h = 24 * 1.6 - 11/2 * 9.81 * 1.6²

h = 38.4 - 12.55 = 25.85 m

since height above the ground is 1.2 m,

total height h = 25.85 m + 1.2 m

h = 27.05 m

4. Let the time of travel of the red ball be t seconds.

So the time of travel of the blue ball = (t - 0.4) seconds.

Both the balls are at the same height :

25 - s = 1.2 + h  where s & h are the displacements of the red & the blue ball respectively.

25 - (ut + 1/2 gt2) = 1.2 + (ut - 1/2 gt2)

25 - (1.6 t + 0.5 * 9.8 t²) = 1.2 + (24(t-0.4) - 0.5*9.8*(t-0.4)²)

solving the equation above for the time after which both the balls are at the same height.

25 - 1.6t - 4.9t² = 1.2 + 24t - 9.6 - 4.9t² + 3.92t - 0.784

collecting like terms

(25 - 1.2 + 9.6 + 0.784) = (24 + 3.92 + 1.6) * t  

t = 34.184 / 33.44

t = 1.16 seconds

Complete and Clear Question:

A person puts a few apples into the freezer at -15°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be [tex]\rho =[/tex] 840 kg/m3,  [tex]c_{p} =[/tex] 3.81 kJ/kg·K, k = 0.418 W/m·K, and [tex]\alpha = 1.3 * 10^{-7} m^{2} /s[/tex], determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Answer:

Temperature at the center of the apple, T(t) = 11.215°C

Temperature at the surface of the apple, T(r,t) = 2.68°C

Amount of heat transfer from each apple, Q = 21.47 kJ

Explanation:

For clarity and easiness of expression, the calculations are handwritten and attached as a file. Check the attached files for the complete calculation.

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

ΔP = 0.056 psi

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{5.2}{0.35}} \\\\f=0.613\ Hz[/tex]

Time period:

[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.613}\\\\T=1.63\ s[/tex]

Maximum velocity in the spring is given by :

[tex]v=A\omega[/tex]

[tex]v=A\sqrt{\dfrac{k}{m}} \\\\v=0.1\times \sqrt{\dfrac{5.2}{0.35}} \\\\v=0.38\ m/s[/tex]

The maximum force acting in the spring is :

[tex]F=-kx\\\\F=kA\\\\F=5.2\times 0.1\\\\F=0.52\ N[/tex]

Hence, this is the required solution.

Answer:

176.38 rpm

Explanation:

mass percentage of arms and legs = 13%

mass percentage of legs and trunk = 80%

mass percentage of head = 7%

Total mass of the skater = 74.0 kg

length of arms = 70 cm = 0.7 m

height of skater = 1.8 m

diameter of trunk = 35 cm = 0.35 m

Initial angular momentum = 68 rpm

We assume:

We split her body into two systems, the spinning hands as spinning rods

1. Each rod has moment of inertia = [tex]\frac{1}{3} mL^{2}[/tex]

mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg

mass of each side will be assumed to be 9.62/2 = 4.81 kg

L = length of each arm

therefore,

I =  [tex]\frac{1}{3}[/tex] x 4.81 x [tex]0.7^{2}[/tex] = 0.79 kg-m   for each arm

2. Her body as a cylinder has moment of inertia =  [tex]\frac{1}{2} mr^{2}[/tex]

r = radius of her body = diameter/2 = 0.35/2 = 0.175 m

mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg

I = [tex]\frac{1}{2}[/tex] x 64.38 x [tex]0.175^{2}[/tex] = 0.99 kg-m

We consider each case

case 1: Body spinning with arm outstretched

Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk

I = (0.79 x 2) +  0.99 = 2.57 kg-m

angular momentum = Iω

where ω = angular speed = 68.0 rpm = [tex]\frac{2\pi }{60}[/tex] x 68 = 7.12 rad/s

angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s

case 2: Arms pulled down parallel to trunk

The momentum of inertia will be due to her body trunk alone which is 0.91 kg-m

angular momentum =  Iω

=  0.99 x ω = 0.91ω

according to conservation of angular momentum, both angular momentum must be equal, therefore,

18.29 = 0.99ω

ω = 18.29/0.99 = 18.47 rad/s

18.47 ÷ [tex]\frac{2\pi }{60}[/tex]  = 176.38 rpm

Answer: the answer is MIRANDA COOPER

Explanation:

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Answer:

F = 1.2×10⁻³ N

Explanation:

From the question,

Applying newton's second law of motion,

F = m(v-u)/t................... Equation 1

Given: F = magnitude of the average force exerted on the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time of  contact.

Note: let downward be negative and upward be positive.

Given: m = 48 g = 48/1000 = 0.048 kg, v = 17 m/s, u = -28 m/s (downward),

t = 1800 s

Substitute into equation 1

F = 0.048(17-[28])/1800

F = 1.2×10⁻³ N

Answer:

The image is virtual

number-4

Answer:

[tex]c_B=0.0768\frac{kcal}{kg\cdot\°C}[/tex]

Explanation:

In order to calculate the specific heat of the material B, you use the following formula for the change in the temperature of a substance, where an amount of heat Q is given to the substance:

[tex]Q=mc(T_2-T_1)[/tex]

Q: amount oh heat

m: mass of the substance

T2: final temperature

T1: initial temperature

c: specific heat of the substance.

If QA and QB are the heat of material A and B, you have:

[tex]Q_A=m_Ac_A(T_{2A}-T_{1A})\\\\Q_B=m_Bc_B(T_{2B}-T_{1B})[/tex]

both materials have the same mass, mA = mB

cA: specific heat of A = 0.2007 kcal/(kg.°C)

cB: specific heat of B = ?

T2A: final temperature of A = 86.3°C

T1A: initial temperature of A = 22.0°C

T2B: final temperature of B = 190.0°C

T1B: initial temperature of B = 22.0°C

In this case you have that both material A and B receive the same amount of heat Q. Then, you equal QA with QB and solve for cB:

[tex]m_Ac_A(T_{2A}-T_{1A})=m_Bc_B(T_{2B}-T_{1B})\\\\c_B=\frac{c_A(T_{2A}-T_{1A})}{(T_{2B}-T_{1B})}\\\\c_B=\frac{(0.2007kcal/(kg.\°C))(86.3\°C-22.0\°C)}{190.0\°C-22.0\°C}\\\\c_B=0.0768\frac{kcal}{kg\cdot\°C}[/tex]

hence, the specific heat of the second rod B is 0.0768kcal/(kg°C)

Answer:

as centripatal force acts upon an object moving at a circle in constant speed de force acts always inwards as de velocity of object is directed tangent to de circle de force can accelerate de object by changing its direction bt not actually de speed

The two quantities are closely related, but the cause/effect is the other way around.

-- The centripetal force is caused by something outside this discussion, not by the object.

-- The centripetal force acting on the object determines the object's centripetal acceleration.

-- The centripetal acceleration is the cause of whatever the object's velocity (speed and direction) turns out to be.

-- It's the centripetal acceleration that has the effect on the object's velocity.  

As an example, you wouldn't say that the orbiting of a TV satellite is what causes the Earth's centripetal force that acts on it.

Answer:

Step 1

Work done = -9134.4 J

ΔQ = -9134.4 J

Step 2

ΔQ =  -3570.32 J = ΔU

W = 0

Step 3

The pdV work done = 3570.32 J

The Vdp work done = 11053.37 J

Heat transferred, ΔE = 0.

Explanation:

For diatomic gases γ = 1.4

Step 1

Where:

v₂ = 3.00·v₁

On isothermal expansion of an ideal gas by Boyle's law, we have;

p₁·v₁ = p₂·v₂ which gives;

p₁·v₁ = p₂×3·v₁

Dividing both sides by v₁, we have;

p₁= 3·p₂

[tex]p_2 = \dfrac{p_1}{3}[/tex]

Hence, the pressure is reduced by a factor of 3

Work done =

[tex]n\cdot R\cdot T\cdot ln\dfrac{v_{f}}{v_{i}}[/tex]

Where:

n = 1 mole  

R = 8.3145 J/(mole·K)

T = 1000 K we have

[tex]1 \times 8.3145 \times 1000 \times ln\left (\dfrac{1}{3} \right ) = -9134.4 J[/tex]

Step 2

The gas undergoes a constant volume decrease in pressure given by Charles law as follows;

[tex]\dfrac{p_2}{p3} = \dfrac{T_1}{T_3}[/tex]

Whereby p₂ > p₃, T₁ will be larger than T₃

W = 0 for constant volume process

ΔQ = m×cv×ΔT = 1 × 3.97 × -900 = -3570.32 J = ΔU

Step 3

For adiabatic compression, we have;

[tex]\dfrac{p_3}{p_1} = \left (\dfrac{V_1}{V_3} \right )^{\gamma } = \left (\dfrac{T_3}{T_1} \right )^{\frac{\gamma }{\gamma -1}}[/tex]

Where:

T₁ = 1000 K

T₃ = 100 K

We have;

[tex]\left (\dfrac{V_1}{3\cdot V_1} \right )^{\gamma } = \left (\dfrac{100}{1000} \right )^{\dfrac{\gamma}{\gamma -1}}[/tex]

[tex]\left (\dfrac{1}{3} \right ) = \left (\dfrac{1}{10} \right )^{\dfrac{1}{\gamma -1}}[/tex]

[tex]log\left (\dfrac{1}{3} \right ) = {\dfrac{1}{\gamma -1}} \times log \left (\dfrac{1}{10} \right )^[/tex]

[tex]\gamma -1 =\dfrac{log \left (\dfrac{1}{10} \right )}{ log\left (\dfrac{1}{3} \right ) } {[/tex]

∴ γ-1 = 2.096

γ = 3.096

The pdV work done =

[tex]m \times c_v \times (T_1 - T_3)[/tex]

m×R/(γ - 1)×(T₁ - T₃) =

3.97×(1000 - 100) = 3570.32 J

The Vdp work done =

[tex]m \times c_p \times (T_1 - T_3)[/tex]

[tex]c_p = k \times c_v = 3.096 \times 3.97 = 12.3 \, J/(mol\cdot K)[/tex]

12.3×(1000 - 100) = 11053.37 J

Heat transferred, ΔE = 0.