Answer:
(a) The number of gallons of water in the tank when it starts draining.
(b) How many minutes it takes for all of the water to drain from the tank.
Explanation:
So, basically v(t) defines the volume of tank as a function of time. When we evaluate t=0, we find the intercept with the y-axis which represent the initial water volume of the tank in gallons:
now we need to know when v(t)=0, so:
Solving for t:
This represents the amount of time in minutes required to drain all of the water from the tank, in other words v=0gal
I attached a picture where you can see the graph that represents the function:
Answer:
Q= - 7 KJ
Heat is rejected
Explanation:
In first process:
Process is adiabatic so heat transfer will be zero.
Q= 0 For adiabatic process
Now from first law of thermodynamics
Q = ΔU + W
ΔU is the change in internal energy
Q is the heat transfer
W is the work.
So here Q= 0
And work is transfer to the system .It means that work done on the system so it will be taken as negative.
W= - 12 KJ
Q = ΔU + W
0 = ΔU -12
ΔU = 12 KJ
It means that in first process internal energy of system increase.
In second process:
non-adiabatic process and work done by system is 5 KJ.
Q = ΔU + W
U is the point function and it does not depends on the path follows it depends only on final and initial states.
So fro second process
ΔU = - 12 KJ
Q= -12 + 5
Q= - 7 KJ
In magnitude Q= 7 KJ
It mean that heat is rejected from the system because it give negative sign.
Answer:
The capacity of the sludge pump is 0.217 m3/min
Explanation:
Solution is attached below
Answer:49.3
Explanation:
Given
mass of water 
mass of tank 
initial of temperature of water =
initial of temperature of tank=
Specific heat of water =4.184kJ/kg k
Specific heat of copper=0.386 KJ/kg k
From first law of thermodynamics we have
Given tank is insulated thus Q=0
work will be negative as it is being done on system
