Question
Asked Dec 3, 2019
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A closed, rigid tank contains 4 kg of air initially at 300 K, 1 bar. As illustrated in the figure below, the tank is in contact with a thermal reservoir at 600 K and heat transfer occurs at the boundary where the temperature is 600 K. A stirring rod transfers W = -480 kJ of energy as shown in the diagram.The final temperature is 600 K. The air can be modeled as an ideal gas with cν = 0.733 kJ/kg · K and kinetic and potential energy effects are negligible.

Determine the amount of entropy transferred into the air and the amount of entropy produced, each in kJ/K.


Q/Tb =  kJ/K
σ =  kJ/K
This portion of the
boundary is at temperature Tres 600 K
Reservoir
at Tres
W
- Air initially at 300 K, 1 bar.
Finally, T2 600 K
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This portion of the boundary is at temperature Tres 600 K Reservoir at Tres W - Air initially at 300 K, 1 bar. Finally, T2 600 K

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Expert Answer

Step 1

Consider ΔU as the change in internal energy, Q as the heat, and W as the work. The kinetic energy (K.E.) and potential energy (P.E.) are neglected. So,

AK.E.= 0
AP.E. 0
Thus, from the energy balance equation,
AK.E.+AP.E.AU = Q -W
0+0+AU = Q-W
AU Q-W
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AK.E.= 0 AP.E. 0 Thus, from the energy balance equation, AK.E.+AP.E.AU = Q -W 0+0+AU = Q-W AU Q-W

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Step 2

Now, consider m as the mass, CV as the specific heat at constant volume, T2 and T1 as the final and initial temperature respectively. Thus, from the above equation, calculate the value of Q.

тC, (T, —7) - 0-w
о- тC, (T, —Т)+w
4 kg x 0.733 kJ/kg K ( 600 K - 300 K)-480 kJ
=399.6 kJ
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тC, (T, —7) - 0-w о- тC, (T, —Т)+w 4 kg x 0.733 kJ/kg K ( 600 K - 300 K)-480 kJ =399.6 kJ

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Step 3

Consider Tb as the boundary temperature. Calculate the...

о
Т,
399.6 kJ
600 K
0.666 kJ/K
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о Т, 399.6 kJ 600 K 0.666 kJ/K

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