Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density

Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu
Answer: The exit temperature of the gas in deg C is
.
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa, 

We know that for an ideal gas the mass flow rate will be calculated as follows.

or, m = 
=
= 10 kg/s
Now, according to the steady flow energy equation:




= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
= 
Therefore, we can conclude that the exit temperature of the gas in deg C is
.
Answer:
16
Explanation:
Have a good day my friend
Answer:
Percentage of solution=32. 96%
Explanation:
Given that initially tank have 50% solution.It means that amount of solution=0.5 x 2500 =1250 lts
Lets take the amount of solution at time t = A
So




Now by integration

Where C is the constant
Given that at t=0 ,A=1250
So 

When t=20 min

A=837.90
So percentage of solution after 20 min

Percentage of solution=32. 96%