A tank initially contains 130 liters of water with 45 grams of salt. The tank is
ID: 2871946 • Letter: A
Question
A tank initially contains 130 liters of water with 45 grams of salt.
The tank is rinsed with fresh water flowing in at a rate of 5 liters per minute
and leaving the tank at the same rate. The water in the tank is well-stirred.
a.(1/10) Find the function V volume of water in the tank, in liters,
b. (3/10) Find the differential equation satisfied by the function values Q(t),
the total amount of salt in the tank at time t, measured in grams,
Q'(t)=
c. (4/10) Find the function Q, the amount of salt grams in the tank as function of time,
Q(t)=
d. (2/10) Find the time t such that the amount of salt in the tank is 20 grams,
t=
Explanation / Answer
a)volume of water in tank =130+(5-5)*t =130 litres
b)r in rate of salt poured into the tank per min. Since it is fresh water, so it is zero=0 , r out rate of salt flowing out=(Q/130)*5 =Q/26 ,Q(0) =45gm
Q'(t)=rin -r out =-Q/26
c)dQ/dt=-Q/26 ==>dQ/Q=-dt/26 ==>integrate ==>integraldQ/Q=-integraldt/26 ==>lnQ=-t/26 +c==>Q(t)=Ce-t/26
Q(0)=45==>45=Ce-0/26==>45=C*1==>C=45 ==>Q(t)=45e-t/26
d) time t such that the amount of salt in the tank is 20 grams
20=45e-t/26
==>e-t/26 =20/45
==>t=21.1
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