Respuesta :
Answer:
a) the final temperature of the air is 500° R
b) the amount of specific entropy produced is 0.0758 Btu/lb-°R
Explanation:
Given the data in the question;
Air at 500° R = [tex]T_i[/tex]
Using first law of thermodynamic;
δQ = dU + W
now, since the vessel is insulated, the transfer is zero, work done also is zero since there is also no external work done.
δQ = dU + W
0 = dU + 0
dU = 0
[tex]u_f[/tex] - [tex]u_i[/tex] = 0
[tex]u_f[/tex] = [tex]u_i[/tex]
hence, change in internal energy is 0
Now, since the ideal internal energy is a function of temperature, the temperature will also remain the same;
[tex]T_f = T_i[/tex]
F = 500° R
Therefore, the final temperature of the air is 500° R
b)
given that; initial volume is one-third of the total volume
V₁ = [tex]\frac{1}{3}[/tex]V₂
3V₁ = V₂
3 = V₂/V₁
Now, we take the value of gas constant R from air property table; gas constant R = 0.069 Btu/lb-R
so we calculate the entropy change;
Δs = [tex]c_v[/tex]In( [tex]\frac{T_2}{T_1}[/tex] ) + R.In( [tex]\frac{V_2}{V_1}[/tex] )
we substitute
Δs = [tex]c_v[/tex]In( [tex]\frac{500}{500}[/tex] ) + 0.069 × In( 3 )
Δs = 0 + [0.069 × In( 3 )]
Δs = 0 + [0.069 × 1.0986]
Δs = 0.0758 Btu/lb-°R
Therefore, the amount of specific entropy produced is 0.0758 Btu/lb-°R
Insulated vessels separate the environment of the outer and the inner system. The final temperature is 500 degrees R and 0.0758 Btu/lb- degree R is the entropy.
What is temperature?
The temperature is the measure of the hot or the coldness of the system. The first law of the thermodynamics is used to measure the final temperature of the system:
[tex]\rm \Delta Q = \rm \Delta U + W[/tex]
The work done will be zero as the system is insulated and no external work is being done.
[tex]\begin{aligned} \rm 0 &= \rm \Delta U + 0\\\\\rm U_{f} - U_{i} &= 0\\\\\rm U_{f} &= \rm U_{i} \end{aligned}[/tex]
Hence, the change in the internal energy is zero. Thus, the final temperature will remain the same,
[tex]\rm T_{f} = \rm T_{i} = 500 ^{\circ} \rm R[/tex]
Now, as we know, the initial volume is one-third of the total volume then,
[tex]\begin{aligned} \rm V_{1} &= \rm \dfrac{1}{3} V_{2}\\\\\rm 3V_{1}&= \rm V_{2}\\\\3 &= \rm \dfrac{V_{1}}{V_{2}}\end{aligned}[/tex]
The change in entropy is calculated as:
[tex]\begin{aligned} \rm \Delta S &= \rm C_{v} ln ( \dfrac{T_{2}}{T_{1}}) + R \times ln ( \dfrac{V_{2}}{V_{1}}) \\\\&= 0 + [0.069 \times \rm ln( 3 )]\\\\& = 0.0758 \;\rm Btu/lb-^{\circ}R \end{aligned}[/tex]
Therefore, the entropy produced is 0.0758 Btu/lb- degree R.
Learn more about entropy here:
https://brainly.com/question/19538748
