Respuesta :
Answer:
1.10 quarts
Step-by-step explanation:
We are given that
Capacity of system=20 quarts=[tex]\frac{20}{4}=5 gal[/tex]
1 gallon=4 quarts
Initially mixture contains
Antifreeze=3 quarts=[tex]\frac{3}{4}=0.75 gal[/tex]
Water=17 quarts=[tex]\frac{17}{4}=4.25 gal[/tex]
Water runs in to the system=1gal/min
We have to find the antifreeze in the system at the end of 5 Â minutes.
Concentration inflow=0
Let Concentration outflow=x
Rate of system=Concentration inflow(water inflow rate)-Concentration outflow(water outflow rate)
[tex]\frac{dx}{dt}=0\times 1-\frac{1}{5} x=-\frac{x}{5}[/tex]
[tex]\frac{dx}{x}=-\frac{dt}{5}[/tex]
Taking integration on both sides
[tex]\int\frac{dx}{x}=-\frac{1}{5}\int dt[/tex]
[tex]ln x=-\frac{1}{5}t+C[/tex]
[tex]x=3 and t=0[/tex]
[tex]ln 3=C[/tex]
[tex] ln x-ln 3=-\frac{1}{5}t[/tex]
[tex]ln\frac{x}{3}=-\frac{1}{5}t[/tex]
Using formula
[tex]ln m-ln n=ln\frac{m}{n}[/tex]
[tex]\frac{x}{3}=e^{-0.2t}[/tex]
[tex]x=3e^{-0.2t}[/tex]
Substitute t=5
[tex]x=3e^{-0.2\times 5}=3e^{-1}=1.10[/tex]
Hence, in the end of 5 minute ,the antifreeze in the system=1.10 quarts
After 5 minutes, there will be 0.98 quarts of antifreeze and 19.02 quarts of water in the tank.
Since a car's cooling system has a capacity of 20 quarts, and initially, the system contains a mixture of 3 quarts of antifreeze and 17 quarts of water, and water runs into the system at the rate of 1 gal min, then the homogeneous mixture runs out at the same rate, to determine, in quarts, how much antifreeze is in the system at the end of 5 minutes, the following calculation must be performed:
- 1 quart = 0.25 gallons
- 3 x 0.25 = 0.75 gallons of antifreeze
- 17 x 0.25 = 4.25 gallons of water
- 1 = 0.2 of 5
- Minute 1 = 0.75 x 0.8 = 0.6
- Minute 2 = 0.6 x 0.8 = 0.48
- Minute 3 = 0.48 x 0.8 = 0.384
- Minute 4 = 0.384 x 0.8 = 0.3072
- Minute 5 = 0.3072 x 0.8 = 0.24576
- 0.24576 gallons = 0.98 quarts
Therefore, after 5 minutes, there will be 0.98 quarts of antifreeze and 19.02 quarts of water in the tank.
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