= Heat released to cold reservoir
= Heat released to hot reservoir
= maximum amount of work
= temperature of cold reservoir
= temperature of hot reservoir
we know that
eq-1
maximum work is given as
=
-
using eq-1
=
-
Answer:
v_{f} = 115.95 m / s
Explanation:
This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions
Thrust =
-v₀ = v_{e}
where v_{e} is the velocity of the gases relative to the rocket
let's apply these expressions to our case
the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units
M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg
The final mass is the mass of the engines + the mass of the rocket
M_{f} = 25.5 +54.5 = 80 g = 0.080 kg
thrust and duration of ignition are given
thrust = 5.26 N
t = 1.90 s
Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear
thrust = v_{e}
v_{e} = thrust
v_{e} = 5.26
v_{e} = - 786.93 m / s
the negative sign indicates that the direction of the gases is opposite to the direction of the rocket
now we look for the final speed of the rocket, which as part of rest its initial speed is zero
v_{f}-0 = v_{e}
we calculate
v_{f} = 786.93 ln (0.0927 / 0.080)
v_{f} = 115.95 m / s