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Anna007 [38]
2 years ago
7

A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer

red out of the gas. part a what is the change in thermal energy of the gas during this process?
Physics
1 answer:
Mekhanik [1.2K]2 years ago
7 0
The initial volume of the gas is
V_i = 600 cm^3
while its final volume is
V_f = 200 cm^3
so its variation of volume is
\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3

The pressure is constant, and it is
p=400 kPa = 400 \cdot 10^3 Pa

Therefore the work done by the gas is
W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
Q=+100 J

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
\Delta U = Q-W = 100 J - (-160 J)=+260 J
where the positive sign means the internal energy of the gas has increased.
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Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1
7nadin3 [17]

Answer:

A). σ = 3.823 x 10^{-5} C^{2}/N-m^{2}

B). \sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C). U=10.322 J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

\sigma =k.\varepsilon _{0}.E

where, E is resultant electric field = 1.2 x 10^{6} V/m

           \varepsilon _{0} is permittivity of free space = 8.85 x 10^{-12} C^{2}/N-m^{2}

           k is dielectric constant = 3.6

∴\sigma =k.\varepsilon _{0}.E

                     = 3.6 x 8.85 x10^{-12} x 1.2 x 10^{6}

                    = 3.823 x 10^{-5} C^{2}/N-m^{2}

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )

\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )

\sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C).

Area of the plate, A = 2.5 cm^{2}

                                 = 2.5 x 10^{-4}m^{2}

diameter of the plate, d = 1.8 mm

                                        = 1800 m

∴ Total energy stored in the capacitor

U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad

U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800

U=10.322 J

4 0
2 years ago
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
2 years ago
5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by
Anna007 [38]

Answer:

a) W=2.425kJ

b) \Delta E=2.425kJ

c) T_f=20.06^{o}C

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

W=Fd

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

W=mgh

so now e can substitute:

W=(25kg)(9.7 m/s^{2})(10m)

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

\Delta E=2.425kJ

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

\Delta Q=mC_{p}(T_{f}-T_{0})

Where:

Q= heat transferred

m=mass

C_{p}=specific heat

T_{f}= Final temperature.

T_{0}= initial temperature.

So we can solve the forula for the final temperature so we get:

T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}

So now we can substitute the data we know:

T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C

Which yields:

T_{f}=20.06^{o}C

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

\Delta Q=-2.425kJ

3 0
2 years ago
Unlike acceleration and velocity, speed does not need to specify
frosja888 [35]

Unlike acceleration and velocity, speed does not need to specify the direction of motion. Speed is a scalar quality.

4 0
1 year ago
Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo
nalin [4]

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

3 0
2 years ago
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