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forsale [732]
3 years ago
14

An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1000 K and a heat rejection at 400 K. During t

he heat addition the volume triples. Find the two specific heat transfers (q) in the cycle and the overall cycle efficiency
Physics
1 answer:
vivado [14]3 years ago
6 0

Answer:

W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600

Explanation:

The Carnot cycle is described by

      e= 1 - Q_{c} / Q_{H} = 1 - T_{c} / T_{H}

     

In this case they indicate that the final volume is

         V = 3V₀

In the part of the heat absorption cycle from the source is an isothermal expansion

         W = n RT ln (V₀ / V)

         W / n = 8.314 1000 ln (1/3)

          W / n = - 9133 J / mol

During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times

            W / n = 8.314 400 (3)

           W / n = 3653 J / mol

The efficiency of the cycle is

            e = 1- 400/1000

            e = 0.600

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7 0
1 year ago
The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?
RoseWind [281]

Answer:

E = 2.9775\times10^5 J

Explanation:

Given:  The latent heat of fusion for Aluminum is L = 3.97\times10^5  J/Kg

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J

Therefore, energy required to melt 0.75 Kg aluminum

E = 2.9775\times10^5 J

5 0
2 years ago
What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are
nexus9112 [7]

Answer:

Density of 18.0-karat gold mixture is 15.58 g/cm^3.

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = 19.32g/cm^3

The density of silver = 10.1g/cm^3

The density of copper =8.8g/cm^3

Volume=\frac{Mass}{Density}

Volume of the gold in the mixture = V_1=\frac{18x}{19.32 g/cm^3}

Volume of the silver in the mixture = V_2=\frac{5x}{10.1 g/cm^3}

Volume of the copper in the mixture = V_3=\frac{1x}{8.8 g/cm^3}

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = V_1+V_2+V_3

Density of the mixture:

\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3

8 0
3 years ago
A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph
posledela

light and radiation is able to experience by an observer watching the explosion from the vacuum.

<u>Explanation:</u>

  • Basically in the nuclear explosion, there will be an enormous of energy released as noise, heat, visible light, radiation and atmospheric wave.
  • Usually sound and wave propagate through medium but in vacuum, there will be no medium to transfer this type of energy. so there is no chance of sound and wave transfer.  
  • Light and radiation travel in vacuum because they didn't need the medium to transfer .  

7 0
2 years ago
Read 2 more answers
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Alexeev081 [22]
You might want to do a re-erp on this equation
5 0
3 years ago
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