Answer:
a)P₂ =4 bar
b)W= - 1482.48 KJ
It means that work done on the system.
c)S₂ - S₁ = 3.42 KJ/K
Explanation:
Given that
T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar
T₂ = 600 K ,V₂=V₁ 3 m³
Given that tank is rigid and insulated.It means that volume of the gas will remain constant.
Lets take the final pressure = P₂
For ideal gas P V = m R T
P₂ =4 bar
Internal energy
ΔU = m Cv ΔT
Cv=0.71 KJ/kg.k for air
m= 6.96 kg
ΔU= 6.96 x 0.71 x (600 - 300)
ΔU=1482.48 KJ
From first law
Q= ΔU + W
Q= 0 Insulated
W = - ΔU
W= - 1482.48 KJ
It means that work done on the system.
Change in the entropy
S₂ - S₁ = 3.42 KJ/K
Answer: ask other people if they like it, or ask what they want to add
Explanation: because maybe other people can can help him improve and brainstorm with other’s.
Heat energy and thermal energy are the same because heat energy is thermal energy. Also thermal energy and temperature are the same because temperature is measuring heat in degrees Celsius or degrees Fahrenheit. Hope this helps!
Answer:
0.9432 m/s
Explanation:
We are given;
Mass of swimmer;m_s = 64.38 kg
Mass of log; m_l = 237 kg
Velocity of swimmer; v_s = 3.472 m/s
Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.
So;
Initial momentum = final momentum
m_l × v_l = m_s × v_s
Where v_l is speed of the log relative to water
Making v_l the subject, we have;
v_l = (m_s × v_s)/m_l
Plugging in the relevant values, we have;
v_l = (64.38 × 3.472)/237
v_l = 0.9432 m/s
