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atroni [7]
3 years ago
13

Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1

970 J of thermal energy into a volume of 47.9 cm3 that is at a gas pressure of 1.18×106Pa. In this short period of time the volume of the gas can be considered constant.
Treating the fuel-air mixture as a monatomic ideal gas at an initial temperature of 325 K, what is its temperature after ignition?
Physics
2 answers:
Tju [1.3M]3 years ago
5 0

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

JulsSmile [24]3 years ago
5 0

Answer:

The temperature after ignition is 7859.565 K

Explanation:

Here we have

Energy released = 1970 J

Volume of gas = 47.9 cm³ = 0.0000479 m³

Pressure of gas = 1.18 × 10⁶ Pa

Temperature of he gas = 325 K

From P·V = n·R·T  

Therefore, n = PV/(RT) = (0.0000479 × 1.18 × 10⁶)/(8.3145× 325) = 2.09 × 10⁻² moles    

For an ideal mono-atomic gas, the mono-atomic gas, the molar heat capacity at constant volume is given as

C_v = 3/2 R ≈ 12.5 J K−1 mol⁻¹

Therefore we have

Heat released, ΔH = 1970 J = n × C_v×(T₂ - T₁) = 2.09 × 10⁻² × 12.5×(T₂ -325)

T₂ = (1970 J)/(2.09 × 10⁻² × 12.5) + 325  

    = 7534.565 +325 = 7859.565 K

The temperature after ignition = 7859.565 K.

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A block of ice at 0°C is added to a 150g aluminum calorimeter cup which holds 210 g of water at 12°C. If all but 2.0 g of ice me
Over [174]

Answer:

original mass of the block of ice is 38.34 gram

Explanation:

Given data

cup mass = 150 g

ice temperature = 0°C

water mass = 210 g

water temperature = 12°C

ice melt = 2 gram

to find out

solution

we know here

specific heat of aluminum is c = 0.900 joule/gram °C

Specific heat of water C =  4.186 joule/gram °C

so here temperature difference is dt =  12- 0 = 12°C

so here heat lost by water and cup are given by

heat lost  = cup mass × c  × dt + water  mass × C × dt

heat lost  = 150 × 0.900  × 12 + 210 × 4.186 × 12

heat lost  = 12168.72 J

so

mass of ice melt here = heat lost / latent heat of fusion

here we know latent heat of fusion = 334.88 joule/gram

so

mass of ice melt  =  12168.72 / 334.88

mass of ice melt  is 36.337554 gram

so mass of ice is here = mass of ice melt + ice melt

mass of ice  =  36.337554 + 2

mass of ice  =  38.337554 gram

so original mass of the block of ice is 38.34 gram

7 0
3 years ago
How does quantum mechanics relate to the universe?
Rom4ik [11]
It is the description of all the microscopic objects in everything in the univverse
4 0
3 years ago
The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
pochemuha

Since g is constant,  the force the escaping gas exerts on the rocket will be 10.4 N

<h3>What is Escape Velocity ?</h3>

This is the minimum velocity required for an object to just escape the gravitational influence of an astronomical body.

Given that the velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in 0.60s. The gravitational field intensity is 9.8N/kg.

To calculate the force the escaping gas exerts of the rocket, let first highlight all the given parameters

  • Mass (m) of the rocket 0.25 Kg
  • Initial velocity u = 15 m/s
  • Final Velocity v = 40 m/s
  • Time t = 0.6s
  • Gravitational field intensity g = 9.8N/kg

The force the gas exerts of the rocket = The force on the rocket

The rate change in momentum of the rocket = force applied

F = ma

F = m(v - u)/t

F = 0.25 x (40 - 15)/0.6

F = 0.25 x 41.667

F = 10.42 N

Since g is constant,  the force the escaping gas exerts on the rocket is therefore 10.4 N approximately.

Learn more about Escape Velocity here: brainly.com/question/13726115

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7 0
1 year ago
A 2.0 kg box is placed on an inclined plane that forms an angle of 21 ° with the horizontal
Dmitrij [34]

Answer:

.-2.0

Explanation:

as at 80 friction it's a 360?

5 0
3 years ago
A package is dropped from an airplane flying horizontally with constant speed V in the positive xdirection. The package is relea
k0ka [10]

Answer:

D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}

Explanation:

From the vertical movement, we know that initial speed is 0, and initial height is H, so:

Y_{f}=Y_{o}-g*\frac{t^{2}}{2}

0=H-g*\frac{t^{2}}{2}    solving for t:

t=\sqrt{\frac{2H}{g} }

Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:

X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}   Replacing values:

D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}

Simplifying:

D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}

4 0
3 years ago
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