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A reversible cyclic device does work while exchanging heat with three constant temperature reservoirs. The three reservoirs (1,2, and 3) are at temperatures of 1000K, 300K, and 500K. Four hundred kilojoules of heat are transferred from reservoir 1 to the device, and the total work done is 100 kJ. Find the magnitude and direction of the heat transfer from the other two reservoirs.

Answer :

Answer:

Lets take [tex]Q_2[/tex] heat transfer take place from 500 K reservoir to device and  [tex]Q_3[/tex] from device to  300 K reservoir.

From the energy conservation we can say that

[tex]400+Q_2=100+Q_3[/tex]  

[tex]300=Q_3-Q_2[/tex]   -----1

For reversible process

[tex]\dfrac{400}{1000}+\dfrac{Q_2}{500}-\dfrac{Q_3}{300}=0[/tex]  

[tex]5Q_3-3Q_2=600[/tex]        ----2

By solving above two equation

[tex]Q_3=-150 KJ,Q_2=-450KJ[/tex]

But here sign come negative it means that

[tex]Q_2[/tex] heat transfer take place from device to 500 K reservoir and [tex]Q_3[/tex] from 300 K reservoir to device.

                                                                                                   

${teks-lihat-gambar} Netta00
lhmarianateixeira

In this exercise we have to use the knowledge of heat transfer to calculate the heat transferred to the other two reservoirs will be:

The magnitude of the Q2 is -450 in the out direction while the Q3 is -150 in the inward direction.

From the information given in the statement, we have that:

  • Temperatures of 1000K, 300K, and 500K.
  • Four hundred kilojoules of heat are transferred
  • Total work done is 100 kJ.

knowing that from conservation we can say:

[tex]400+Q_2=100+Q_3\\300=Q_3-Q_2\\\frac{400}{1000}+\frac{Q_2}{500}-\frac{Q_3}{300}=0\\[/tex]

So solving we have:

[tex]Q_3=-150KJ\\Q_2=-450KJ[/tex]

See more about heat transfer at brainly.com/question/12107378

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