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:
Workdone = 465766038 Joules.
Explanation:
<u>Given the following data;</u>
Mass = 1167
Initial velocity = 10m/s
Final velocity =28m/s
To find the workdone;
We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.
Mathematically, it is given by the equation;
W = Kf - Ki
W = ½MVf² - ½MVi²
Substituting into the equation
W = ½(1167)*28² - ½(1167)*10²
W = ½ * 1361889* 784 - ½ * 1361889 * 100
W = 533860488 - 68094450
Workdone = 465766038 Joules.
Answer:
The volume of water is 139 mL.
Explanation:
Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

The heat (Q) can be calculated using the following expression:

where,
c is the specific heat of each substance
m is the mass of each substance
ΔT is the difference in temperature for each substance
The mass of coffee is:

Then,

The volume of water is:
