Answer:
a)
b)
c) 0 J/K
d)S= 61.53 J/K
Explanation:
Given that
T₁ = 745 K
T₂ = 101 K
Q= 7190 J
a)
The entropy change of reservoir 745 K
Negative sign because heat is leaving.
b)
The entropy change of reservoir 101 K
c)
The entropy change of the rod will be zero.
d)
The entropy change of the system
S= S₁ + S₂
S = 71.18 - 9.65 J/K
S= 61.53 J/K
Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
(1)
Assuming that the velocity remains constant then
Solving for we get
Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust <em>F</em><em> </em> is
The exhaust rate dm/dt is
Therefore, the velocity at which the exhaust gases exit the engines is