I personally don't (I can't speak for others tho) but I say right twix is better for the memes
v
Convert the given temperatures from celsius to kelvin since we are dealing with gas.
To convert to kelvin, add 273.15 to the temperature in celsius.
T1 = 22 + 273.15 = 295.15 k
T2 = 4 + 273.15 = 277.15 k
V1 = 0.5 L
Let's find the final volume (V2).
To solve for V2 apply Charles Law formula below:
Answer:
Rice
Explanation:
Because I can't control eating lots of rice
Answer:
the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg
Explanation:
To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables
Mathematically this can be determined as

Where
Temperature at inlet of turbine
Temperature at exit of turbine
Pressure at exit of turbine
Pressure at exit of turbine
The steady flow Energy equation for an open system is given as follows:

Where,
m = mass
m(i) = mass at inlet
m(o)= Mass at outlet
h(i)= Enthalpy at inlet
h(o)= Enthalpy at outlet
W = Work done
Q = Heat transferred
v(i) = Velocity at inlet
v(o)= Velocity at outlet
Z(i)= Height at inlet
Z(o)= Height at outlet
For the insulated system with neglecting kinetic and potential energy effects

Using the relation T-P we can find the final temperature:


From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg
The speed of an object at rest is 0 m/s. We can use logic to figure this out. Hope this helps!