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notsponge [240]

2 years ago

12

An insulated piston–cylinder device contains 100 L of air at 400 kPa and 25°C. A paddle wheel within the cylinder is rotated unt

il 15 kJ of work is done on the air while the pressure is held constant. Determine the final temperature of the air. Neglect the energy stored in the paddle wheel. (Page 172).

1 answer:

solong [7]2 years ago

7 0

Answer:

T'=70.92°C

Explanation:

Given that

V= 100 L=0.1 m³

P=400 KPa

T=25°C

Work done on the air = 15 KJ

W= -15 KJ

If we assume that air is ideal gas

P V = m R T

R=0.287 KJ/kg.K for air

T= 273 + 25 = 298 K

By putting the values

P V = m R T

400 x 0.1 = m x 0.287 x 298

m=0.46 kg

From first law of thermodynamics

Q= ΔU +W

Insulated piston–cylinder , Q=0

ΔU = m Cv ΔT

ΔU = - W

Cv = 0.71 KJ/kg.k for air

0.46 x 0.71 x (T' -25) = 15

T'=70.92 °C

So the final temperature of air is T'=70.92 °C

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tankabanditka [31]

The correct answer to the question is C i.e C represents the friction from air resistance.

EXPLANATION:

Before coming into any conclusion, first we have to understand friction.

The friction is the opposing force which acts tangentially between two bodies in contact when there is a relative motion between them.

The air resistance is that frictional force which is provided by the air to the moving body through it. Hence, the friction from air resistance will be directed opposite to the motion of the body.

In the given diagram, the airplane is going horizontally. The force A acts in forward direction while force C acts in backward direction. The forces B and D are acting vertically. There is no motion in vertical direction. Hence, the net force of A and C will cause the airplane to move.

As the plane is moving along the direction of A, the frictional force must act along the direction of C.

8 0

2 years ago

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When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same

ehidna [41]

Answer:

7.24 ohm

Explanation:

Let R1 and R2 are resistance of two resistors.

Emf=E=20 V

Current,I=2 A

Current,I'=10 A

We have to find the magnitude of the greater of the two resistances.

In series

$R=R_1+R_2$

$V=IR$

By using the formula

$20=2(R_1+R_2)$

$R_1+R_2=\frac{20}{2}=10$ ...(1)

In parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}$

$R=\frac{R_1R_2}{R_1+R_2}$

$20=10(\frac{R_1R_2}{R_1+R_2}$

$2=\frac{R_1R_2}{10}$

$R_1R_2=20$

$R_2=\frac{20}{R_1}$

Substitute the value

$\frac{20}{R_1}+R_1=10$

$R^2_1+20=10R_1$

$R^2_1-10R_1+20=0$

$R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$R_1=\frac{10\pm 2\sqrt 5}{2}$

$R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm$

$R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm$

Substitute the value

$R_2=\frac{20}{7.24}=2.76 ohm$

$R_2=\frac{20}{2.76}=7.24 ohm$

Hence, the magnitude of the greater of the two resistance=7.24 ohm

8 0

2 years ago

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You exert 500 N of force down on the earth as you jump into the air. How much force does the earth exert back on you?

zalisa [80]

The same 500N, is the Newton’s Third Law.

6 0

2 years ago

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

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3 years ago

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pashok25 [27]

The answer should be D

3 0

2 years ago

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