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dimulka [17.4K]
3 years ago
9

In each cycle, a heat engine performs 710 J of work and exhausts 1480 J of heat. What is the thermal effiency?

Physics
1 answer:
Tomtit [17]3 years ago
6 0

Answer:

32%

Explanation:

For a heat engine, efficiency is work out divided by heat in:

η = Wₒ / Qᵢ

Since energy is balanced, heat in is the sum of work out and heat out:

Qᵢ = Wₒ + Qₒ

Therefore:

η = Wₒ / (Wₒ + Qₒ)

Given Wₒ = 710 J and Qₒ = 1480 J:

η = 710 / (710 + 1480)

η = 0.32

The thermal efficiency is 32%.

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Describe several characteristics that distinguish population I stars from population II stars.
sergij07 [2.7K]

Answer:

Population I stars are usually the blue younger stars. They orbit in the disk and they are relatively rich in metals. Population II stars are the red, older stars that lies in the bulge and halo of the Galaxy. They orbits in the elliptical orbits in the halo.

Population I stars include the sun and tend to be luminous, hot and young, concentrated in the disks of spiral galaxies. They are particularly found in the spiral arms. while, Population II stars tend to be found in globular clusters and the nucleus of a galaxy. They tend to be older, less luminous and cooler than Population I stars. They have fewer heavy elements.

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3 years ago
An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
bixtya [17]

Answer:

a. P=18.61\ W

b. \eta_c=0.6976=69.76\%

c. \eta=0.4718=47.18\%

Explanation:

Given:

  • temperature of the hotter reservoir, T_H=1250\ K
  • temperature of the colder reservoir, T_C=378\ K
  • heat absorbed by the engine, Q_H=1.42\times 10^{5}\ J
  • heat rejected to the cold reservoir, Q_L=7.5\times 10^4\ J
  • time duration of the energy transfer, t=1\ hr=3600\ s

<u>Now the work done by the engine:</u>

Using energy conservation,

W=Q_H-Q_L

W=14.2\times 10^4-7.5\times 10^4

W=6.7\times 10^4\ J

a.

<u>Hence the power output:</u>

P=\frac{W}{t}

P=\frac{6.7\times 10^4}{3600}

P=18.61\ W

b.

\eta_c=1-\frac{T_L}{T_H}

\eta_c=1-\frac{378}{1250}

\eta_c=0.6976=69.76\%

c.

now actual efficiency:

\eta=\frac{W}{Q_H}

\eta=\frac{6.7\times 10^4}{1.42\times 10^{5}}

\eta=0.4718=47.18\%

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Answer:

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Explanation:

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So, six moles of hydrogen chloride (HCl) will react with plenty of aluminum to produce many 2 moles of aluminum chloride (AlCl_3).

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