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Rus_ich [418]
3 years ago
15

A Carnot engine has a power output of 148 kW. The engine operates between two reservoirs at 17◦C and 727◦C. How much thermal ene

rgy is absorbed each hour? Answer in units of J.
Physics
1 answer:
Elanso [62]3 years ago
3 0

Answer:

The thermal Energy absorbed each hour = 750420 kJ

Explanation:

Efficiency of a Carnot engine = W/Q = (Th-Tc)/Th ............... Equation 1.

W/Q = (Th-Tc)/Th

making Q the subject of the equation above,

Q = W(Th)/(Th-Tc)........................ Equation 2

Where Q = input power, W = output power, Th = temperature of the hot reservoir, Tc = Temperature of the cold reservoir.

<em>Given: W= 148 kW, Th = 727 °C = 727 + 273 = 1000 K, Tc = 17  °C =17 + 273 = 290 K.</em>

<em>Substituting these values into equation 2,</em>

<em>Q = 148(1000)/(1000-290)</em>

<em>Q = 148000/710</em>

Q = 208.45 kw

The thermal Energy absorbed each hour = 208.45 kW × 3600 s

The thermal Energy absorbed each hour = 750420 kJ

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TEMPERATURE CHANGES

Explanation:

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What is the solution set of the equation 3|5 -x| + 2 = 29?
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A​ heavy-duty shock absorber is compressed 4 cm from its equilibrium position by a mass of 700nbspkg. How much work is required
tia_tia [17]

Answer:

Explanation:

A mass of 700 kg will exert a force of

700 x 9.8

= 6860 N.

Amount of compression x = 4 cm

= 4 x 10⁻² m

Force constant K = force of compression / compression

= 6860 / 4 x 10⁻²

= 1715 x 10² Nm⁻¹.

Let us take compression of r at any moment

Restoring force by spring

= k r

Force required to compress = kr

Let it is compressed  by small length dr during which force will remain constant.

Work done

dW =  Force x displacement

= -kr -dr

= kr dr

Work done to compress by length d

for it r ranges from 0 to -d

Integrating on both sides

W  = \int\limits^{-4}_0 {kr} \, dr

= [ kr²/2]₀^-4

= 1/2 kX16X10⁻⁴

= .5 x 1715 x 10² x 16 x 10⁻⁴

= 137.20 J

3 0
3 years ago
A 2000 kg truck is traveling at 5 m/s and collides with a 1000 kg car that is not moving. After the collision, the 2000 truck st
sp2606 [1]

Answer:

A) 10 m/s

Explanation:

We know that according to conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2  ..............(equation 1)

where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.

From the given data

If truck and car are two bodies

truck :       m1 = 2000 Kg           v1 = 5 m/s                u1 = 0

car    :        m2 = 1000 kg           v2 = 0                      u2 = ?

final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.

substituting the values in equation 1, we get

(2000 x 5) + 0 = 0 + (1000 x u2)

u2 = \frac{2000}{1000} x 5

    = 10 m/s

Hence after collision, car moves at a velocity of 10 m/s

3 0
3 years ago
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