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Answer :
Final Answer:
It will take approximately 440 seconds for the salt amount in the tank to reduce to 8 kg.
Explanation:
To solve this problem, we will use a differential equation to describe the change in the amount of salt in the tank over time.
Let's denote S(t) as the amount of salt (in kg) in the tank at any time t (in seconds). When the problem begins, S(0) = 112 kg since that is the initial amount of salt in the tank.
Pure water is added to the tank at a rate of 12 L/s, and the same amount of the saltwater mixture is taken out. This means the volume of the solution in the tank remains constant at 2000 L.
The rate at which salt leaves the tank will be proportional to its concentration in the tank at that time, which is [tex]\( \frac{S(t)}{2000} \)[/tex] since the volume of the solution is always 2000 L. Given that the solution is being pumped out at a rate of 12 L/s, the rate at which salt is leaving the tank, [tex]\( \frac{dS}{dt} \)[/tex], will be the concentration times the outflow rate:
[tex]\[ \frac{dS}{dt} = - \left( \frac{S(t)}{2000} \right) \times 12 \][/tex]
This negative sign in this expression indicates that the amount of salt in the tank is decreasing over time.
Let's rearrange this differential equation to separate variables:
[tex]\[ \frac{dS}{S} = - \left( \frac{12}{2000} \right) dt \][/tex]
Now we can integrate both sides to find the relationship between S and t:
[tex]\[ \int \frac{1}{S} dS = -\int \frac{12}{2000} dt \][/tex]
After integrating, we get:
[tex]\[ \ln(|S|) = - \frac{12}{2000} t + C \][/tex]
Where C is the constant of integration. To find C, we use the initial condition S(0) = 112:
[tex]\[ \ln(112) = -\frac{12}{2000} \times 0 + C \\\\\[ C = \ln(112) \][/tex]
Plug C back into the equation for S(t):
[tex]\[ \ln(|S(t)|) = -\frac{12}{2000} t + \ln(112) \][/tex]
We are interested in the time t it takes for the salt quantity S(t) to reach 8 kg:
[tex]\[ \ln(8) = -\frac{12}{2000} t + \ln(112) \][/tex]
Now, solve for t:
[tex]\[ \ln(8) - \ln(112) = -\frac{12}{2000} t \\\\\[ \ln \left( \frac{8}{112} \right) = -\frac{12}{2000} t \\\\\[ \frac{12}{2000} t = -\ln \left( \frac{8}{112} \right) \\\\\[ t = -\frac{2000}{12} \ln \left( \frac{8}{112} \right) \][/tex]
Finally, calculate the value of \( t \):
[tex]\[ t = -\frac{2000}{12} \ln \left( \frac{1}{14} \right) \\\\\[ t \approx -\frac{2000}{12} \ln \left( 0.071428571 \right) \\\\\[ t \approx -\frac{2000}{12} \times (-2.6390573297) \ (using\ a\ calculator\ to\ approximate\ ln(1/14))\\\\\[ t \approx \frac{2000}{12} \times 2.6390573297 \\\\\[ t \approx 166.66667 \times 2.6390573297 \\\\\[ t \approx 439.882 \][/tex]
So it will take approximately 440 seconds for the salt amount in the tank to reduce to 8 kg.
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