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Answer:
Work = 651,1011 kJ
Explanation:
Let´s take the car as a system in order to apply the first law of thermodynamics as follows:
Where
And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:
Regarding the energy system we have the following:
By doing the calculations we have:
![E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ](https://tex.z-dn.net/?f=E_%7Bsystem%2Cfinal%7D-%20E_%7Bsystem%2Cinitial%7D%3D%5B0%2C1%2A900%5D_%7Binternal%7D%2B%5B0%2C5%2A900%2830%5E2-10%5E2%29%2F1000%29_%7Bkinetic%7D%2B%28900%2A10%2A%2820-0%29%2F1000%29_%7Bpotential%7D%5C%5CE_%7Bsystem%2Cfinal%7D-%20E_%7Bsystem%2Cinitial%7D%3D90%2B360%2B180%3D630kJ)
Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.
Finally, the work needed to move the car under the required conditions is calculated as follows:
Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.
Answer:
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Hope this helped you!
Explanation:
Answer: here you go sir
How trains work. On an electric locomotive, the wheels are moved by electric motors. ... On a diesel locomotive, a diesel engine drives the wheels via a mechanical transmission. Cutaway illustration of an electric power carHigh-speed trains are powered by electric current, collected from an overhead cable by a pantograph.
Explanation:
Chugging across short distances or entire continents, trains act as a major form of transportation worldwide. Also called railroads or railways, trains carry within their cars passengers or freight -- such as raw materials, supplies or finished goods -- and sometimes both.
Answer:
The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C
Explanation:
given information:
mass, m = 3 kg
initial temperature, T₁ = 40°C
current, I = 10 A
voltage, V = 50 V
time, t = 30 min = 1800 s
Heat for the system because of the resistance is
Q = V I t
where
V = voltage (V)
I = current (A)
t = time (s)
Q = heat transfer to the system (J)
so,
Q = V x I x t
= 50 x 10 x 1800
= 900000
= 9 x 10⁵ J
the heat transfer in the closed system is
Q = ΔU + W
where
U = internal energy
W = work done by the system
thus,
Q = ΔU + W
9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.
ΔU = 9 x 10⁵ J = 900 kJ
then, the energy change in the system is
ΔU = m c ΔT
ΔT = ΔU / m c, c = 4.186 J/g°C
= 900 / (3 x 4.186)
= 71.67°C
so,the final temperature (T₂)
ΔT = T₂ - T₁
T₂ = ΔT + T₁
= 71.67°C + 40°C
= 111.67°C
