To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.
By definition we know that the change in entropy is given by
![\Delta S = \frac{Q}{T}](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20%5Cfrac%7BQ%7D%7BT%7D)
Where,
Q = Heat transfer
T = Temperature
On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is
![W = Q_{source}-Q_{sink}](https://tex.z-dn.net/?f=W%20%3D%20Q_%7Bsource%7D-Q_%7Bsink%7D)
According to the data given we have to,
![Q_{source} = 200000Btu](https://tex.z-dn.net/?f=Q_%7Bsource%7D%20%3D%20200000Btu)
![T_{source} = 1500R](https://tex.z-dn.net/?f=T_%7Bsource%7D%20%3D%201500R)
![Q_{sink} = 100000Btu](https://tex.z-dn.net/?f=Q_%7Bsink%7D%20%3D%20100000Btu)
![T_{sink} = 600R](https://tex.z-dn.net/?f=T_%7Bsink%7D%20%3D%20600R)
PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is
![\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsink%7D%20%3D%20%5Cfrac%7BQ_%7Bsink%7D%7D%7BT_%7Bsink%7D%7D)
![\Delta S_{sink} = \frac{100000}{600}](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsink%7D%20%3D%20%5Cfrac%7B100000%7D%7B600%7D)
![\Delta S_{sink} = 166.67Btu/R](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsink%7D%20%3D%20166.67Btu%2FR)
On the other hand,
![\Delta S_{source} = \frac{Q_{source}}{T_{source}}](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsource%7D%20%3D%20%5Cfrac%7BQ_%7Bsource%7D%7D%7BT_%7Bsource%7D%7D)
![\Delta S_{source} = \frac{-200000}{1500}](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsource%7D%20%3D%20%5Cfrac%7B-200000%7D%7B1500%7D)
![\Delta S_{source} = -133.33Btu/R](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsource%7D%20%3D%20-133.33Btu%2FR)
The total change of entropy would be,
![S = \Delta S_{source}+\Delta S_{sink}](https://tex.z-dn.net/?f=S%20%3D%20%5CDelta%20S_%7Bsource%7D%2B%5CDelta%20S_%7Bsink%7D)
![S = -133.33+166.67](https://tex.z-dn.net/?f=S%20%3D%20-133.33%2B166.67)
![S = 33.34Btu/R](https://tex.z-dn.net/?f=S%20%3D%2033.34Btu%2FR)
Since
the heat engine is not reversible.
PART B)
Work done by heat engine is given by
![W=Q_{source}-Q_{sink}](https://tex.z-dn.net/?f=W%3DQ_%7Bsource%7D-Q_%7Bsink%7D)
![W = 200000-100000](https://tex.z-dn.net/?f=W%20%3D%20200000-100000)
![W = 100000 Btu](https://tex.z-dn.net/?f=W%20%3D%20100000%20Btu)
Therefore the work in the system is 100000Btu
Explanation:
A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,
![\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%3Dmgh%5C%5C%5C%5Cv%3D%5Csqrt%7B2gh%7D)
h = 2.3 m
![v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%202.3%7D%20%5C%5C%5C%5Cv%3D6.71%5C%20m%2Fs)
So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.
Answer:
0.5kg
Explanation:
Given parameters:
Potential energy = 147J
Height = 30m
Unknown:
Mass of the bird = ?
Solution:
Potential energy is the energy due to the position of a body. Now, the expression for finding the potential energy is given as;
P.E = mgH
m is the mass
g is the acceleration due to gravity = 9.8m/s²
H is the height
147 = m x 9.8 x 30
m = 0.5kg
Answer:
so that each component has the same voltage.
Explanation:
Answer: Take notes and research into the subject more get extra help like tutoring if needed.
Explanation: ...