Respuesta :
Answer:
(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L
Step-by-step explanation:
(a) Initial amount of salt in tank
The tank initially contains 60 kg of salt.
(b) Amount of salt after 4.5 h
[tex]\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}[/tex]
(i) Set up an expression for the rate of change of salt concentration.
[tex]\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}[/tex]
(ii) Integrate the expression
[tex]\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C[/tex]
(iii) Find the constant of integration
[tex]\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)[/tex]
(iv) Solve for A as a function of time.
[tex]\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}[/tex]
(v) Calculate the amount of salt after 4.5 h
a. Convert hours to minutes
[tex]\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}[/tex]
b.Calculate the concentration
[tex]A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}[/tex]
c. Calculate the volume
The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.
The volume added in 4.5 h is
[tex]\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}[/tex]
Total volume in tank = 1000 L + 810 L = 1810 L
d. Calculate the mass of salt in the tank
[tex]\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}[/tex]
(c) Concentration at infinite time
[tex]\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}[/tex]
This makes sense, because the salt is continuously being flushed out by the fresh water coming in.
The graph below shows how the concentration of salt varies with time.
The rate of flow can be described as the amount of fluid flowing past a point in area A per unit of time. The tank initially contains 60 kg of salt.
What is the rate of flow?
The rate of flow can be described as the amount of fluid flowing past a point in area A per unit of time.
A.) Initial amount of salt in the tank
The tank initially contains 60 kg of salt.
B.) Amount of salt 4.5 hours
Let A be the mass of the salt after t min, [tex]r_i[/tex] is the rate of salt coming into the tank, and, [tex]r_o[/tex] is the rate of salt going out of the tank.
1. Set up an expression for the rate of change of salt concentration.
[tex]\dfrac{dA}{dt} = r_i - r_o[/tex]
The freshwater that is entering with no salt, so
[tex]r_i = 0\\\\r_o = \rm{\dfrac{3\ L}{1\ min} \times \dfrac{A\ kg}{1000} = \dfrac{3A}{1000}\ kg/min}\\\\\dfrac{dA}{dt}=-0.003A\ kg/min[/tex]
2. Integrate the expression
[tex]\dfrac{dA}{dt} = -0.003 A\\\\\dfrac{dA}{A}=-0.003dt\\\\\int \dfrac{dA}{A} = -\int0.003dt\\\\\rm ln\ A = -0.003t+C[/tex]
3. Finding the value of the constant of integration
[tex]\rm ln\ A=-0.003t+C\\\\At\ t=0, A=60\ kg/1000\ L = 0.060\ kg/L\\\\ln(0.060)=(-0.003\times 0 )+C\\\\C = ln(0.060)[/tex]
4. Solve for A as a function of time.
The integrated rate expression is
[tex]\rm ln\ A = -0.003t+ln(0.060)\\\\Solve\ for\ A\\\\A = 0.060e^{-0.003t}[/tex]
5. Calculate the amount of salt after 4.5 h
Converting hours to minutes
[tex]\rm Time=4.5\ h \times \dfrac{60\ min}{1\ h} = 270\ min[/tex]
6. Calculate the volume
[tex]A = 0.060e^{-0.003 t}=0.060e^{-0.003 \times 270} = 0.060e^{0.81}=0.060 \times 0.445 = 0.0267\rm\ kg/L[/tex]
C.) Calculate the volume
The tank has been filling at a rate of 6 L/min and draining at 3L/min, so it is filling at a rate of 3 L/min.
The volume added in 4.5 hours is,
[tex]\rm Volume\ added = 270\ min\times \dfrac{3\ L}{1\ min} = 810\ L[/tex]
Total volume in tank = 1000 L+ 810 L = 1810 L
D.) Calculate the mass of salt in the tank
[tex]\rm \text{Mass of salt in tank}=1810\ L \times \dfrac{0.0267\ kg}{1\ L}=21.6\ kg[/tex]
E.) Concentration of salt at infinite time,
[tex]As\ t\rightarrow -\infty ,\ e^{-\infty} \rightarrow 0,\rm\ so\ A\rightarrow 0[/tex]
The reason for this is that the salt is continuously being flushed out by the freshwater coming in.
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