4.2 Cooling and Mixing

Newton’s Law of Cooling

Newton’s law of cooling states that if an object with temperature \(T(t)\) at time \(t\) is in a medium with temperature \(T_m(t)\), the rate of change of \(T\) at time \(t\) is proportional to \(T(t)-T_m(t)\); thus, \(T\) satisfies a differential equation of the form

\begin{equation} T'=-k(T-T_m). \tag{4.2.1}\end{equation}

Here \(k > 0\), since the temperature of the object must decrease if \(T > T_m\), or increase if \(T < T_m\). We’ll call \(k\) the temperature decay constant of the medium.

For simplicity, in this section we’ll assume that the medium is maintained at a constant temperature \(T_m\). This is another example of building a simple mathematical model for a physical phenomenon. Like most mathematical models it has its limitations. For example, it’s reasonable to assume that the temperature of a room remains approximately constant if the cooling object is a cup of coffee, but perhaps not if it’s a huge cauldron of molten metal. (For more on this see Exercise 17.)

To solve (4.2.1), we rewrite it as

\[ T'+kT=kT_m. \]

Since \(e^{-kt}\) is a solution of the complementary equation, the solutions of this equation are of the form \(T=ue^{-kt}\), where \(u'e^{-kt}=kT_m\), so \(u'=kT_me^{kt}\). Hence,

\[ u=T_me^{kt}+c, \]

\[ T=ue^{-kt}=T_m+ce^{-kt}. \]

If \(T(0)=T_0\), setting \(t=0\) here yields \(c=T_0-T_m\), so

\begin{equation} T=T_m+(T_0-T_m)e^{-kt}. \tag{4.2.2}\end{equation}

Note that \(T-T_m\) decays exponentially, with decay constant \(k\).

Example 4.2.1

A ceramic insulator is baked at \(400^\circ\)C and cooled in a room in which the temperature is \(25^\circ\)C. After 4 minutes the temperature of the insulator is \(200^\circ\)C. What is its temperature after 8 minutes?

Solution Here \(T_0=400\) and \(T_m=25\), so (4.2.2) becomes

\begin{equation} T=25+375e^{-kt}. \tag{4.2.3}\end{equation}

We determine \(k\) from the stated condition that \(T(4)=200\); that is,

\[ 200=25+375e^{-4k}; \]

hence,

\[ e^{-4k} = {175\over 375} = {7\over 15}. \]

Taking logarithms and solving for \(k\) yields

\[ k=-{1\over 4} \ln {7\over 15}={1\over 4}\ln {15\over 7}. \]

Substituting this into (4.2.3) yields

\[ T=25+375 e^{-{t\over 4} \ln {15\over 7}} \]

(Figure 4.2.1). Therefore the temperature of the insulator after 8 minutes is

\[\begin{array}{rl} T(8) & \dst = 25+375 e^{-2 \ln {15\over 7}} \\[9pt] & \dst = 25+375 \left({7\over 15}\right)^2 \approx 107^\circ \mbox{C}. \end{array}\]

T=25+375e^-(t/4)15/7 — Figure 4.2.1. \(T=25+375e^{-(t/4)\ln 15/7}\)

Example 4.2.2

An object with temperature \(72^\circ\)F is placed outside, where the temperature is \(-20^\circ\)F. At 11:05 the temperature of the object is \(60^\circ\)F and at 11:07 its temperature is \(50^\circ\)F. At what time was the object placed outside?

Solution Let \(T(t)\) be the temperature of the object at time \(t\). For convenience, we choose the origin \(t_0=0\) of the time scale to be 11:05 so that \(T_0=60\). We must determine the time \(\tau\) when \(T(\tau)=72\). Substituting \(T_0=60\) and \(T_m=-20\) into (4.2.2) yields

\[ T = -20+\bigl(60-(-20)\bigr)e^{-kt} \]

\begin{equation} T = -20+80e^{-kt}. \tag{4.2.4}\end{equation}

We obtain \(k\) from the stated condition that the temperature of the object is 50\(^\circ\)F at 11:07. Since 11:07 is \(t=2\) on our time scale, we can determine \(k\) by substituting \(T=50\) and \(t=2\) into (4.2.4) to obtain

\[ 50 = -20+80e^{-2k} \]

(Figure 4.2.2); hence,

\[ e^{-2k}={70\over 80}={7\over 8}. \]

Taking logarithms and solving for \(k\) yields

\[ k =-{1\over 2} \ln {7\over 8} = {1\over 2} \ln {8\over 7}. \]

Substituting this into (4.2.4) yields

\[ T = -20+80 e^{-{t\over 2}\ln {8\over 7}}, \]

and the condition \(T(\tau)=72\) implies that

\[ 72 =-20+80 e^{-{\tau\over 2} \ln {8\over 7}}; \]

hence,

\[ e^{-{\tau\over 2} \ln {8\over 7}} ={92\over 80} = {23\over 20}. \]

Taking logarithms and solving for \(\tau\) yields

\[ \tau = -{2 \ln {23\over 20}\over \ln {8\over 7}} \approx-2.09\ \mbox{min}. \]

Therefore the object was placed outside about 2 minutes and 5 seconds before 11:05; that is, at 11:02:55.

T = -20+80 e^-t287 — Figure 4.2.2. \(T = -20+80 e^{-{t\over 2}\ln {8\over 7}}\)

Mixing Problems

In the next two examples a saltwater solution with a given concentration (weight of salt per unit volume of solution) is added at a specified rate to a tank that initially contains saltwater with a different concentration. The problem is to determine the quantity of salt in the tank as a function of time. This is an example of a mixing problem. To construct a tractable mathematical model for mixing problems we assume in our examples (and most exercises) that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Exercises 22 and 23 deal with situations where this isn’t so, but the distribution of salt becomes approximately uniform as \(t\to\infty\).

Example 4.2.3

A tank initially contains 40 pounds of salt dissolved in 600 gallons of water. Starting at \(t_0 = 0\), water that contains 1/2 pound of salt per gallon is poured into the tank at the rate of 4 gal/min and the mixture is drained from the tank at the same rate (Figure 4.2.3).

Find a differential equation for the quantity \(Q(t)\) of salt in the tank at time \(t > 0\), and solve the equation to determine \(Q(t)\).
Find \(\lim_{t\to\infty}Q(t)\).

Solution (a) To find a differential equation for \(Q\), we must use the given information to derive an expression for \(Q'\). But \(Q'\) is the rate of change of the quantity of salt in the tank changes with respect to time; thus, if rate in denotes the rate at which salt enters the tank and rate out denotes the rate by which it leaves, then

\begin{equation} Q' = \mbox{rate in}-\mbox{rate out}. \tag{4.2.5}\end{equation}

The rate in is

\[ \left({1\over 2}\ \mbox{lb/gal}\right) \times (4\ \mbox{gal/min}) = 2\ \mbox{lb/min}. \]

Determining the rate out requires a little more thought. We’re removing 4 gallons of the mixture per minute, and there are always 600 gallons in the tank; that is, we’re removing \(1/150\) of the mixture per minute. Since the salt is evenly distributed in the mixture, we are also removing \(1/150\) of the salt per minute. Therefore, if there are \(Q(t)\) pounds of salt in the tank at time \(t\), the rate out at any time \(t\) is \(Q(t)/150\). Alternatively, we can arrive at this conclusion by arguing that

\[ \begin{array}{lcl} \mbox{rate out} & = & (\mbox{concentration})\times(\mbox{rate of flow out})\\[6pt] \mbox{}&=&(\mbox{lb/gal})\times(\mbox{gal/min})\\[10pt] &=&\dst{Q(t)\over600}\times 4=\dst{Q(t)\over150}. \end{array} \]

We can now write (4.2.5) as

\[ Q' = 2-{Q\over 150}. \]

This first order equation can be rewritten as

\[ Q'+{Q\over 150} = 2. \]

Since \(e^{-t/150}\) is a solution of the complementary equation, the solutions of this equation are of the form \(Q=ue^{-t/150}\), where \(u'e^{-t/150}=2\), so \(u'=2e^{t/150}\). Hence,

\[ u = 300e^{t/150}+c, \]

\begin{equation} Q=ue^{-t/150}=300+ce^{-t/150} \tag{4.2.6}\end{equation}

(Figure 4.2.4). Since \(Q(0)=40\), \(c=-260\); therefore,

\[ Q=300-260e^{-t/150}. \]

Solution (b) From (4.2.6), we see that that \(\lim_{t \to \infty}Q(t)=300\) for any value of \(Q(0)\). This is intuitively reasonable, since the incoming solution contains 1/2 pound of salt per gallon and there are always 600 gallons of water in the tank.

Example 4.2.4

A 500-liter tank initially contains 10 g of salt dissolved in 200 liters of water. Starting at \(t_0=0\), water that contains 1/4 g of salt per liter is poured into the tank at the rate of 4 liters/min and the mixture is drained from the tank at the rate of 2 liters/min (Figure 4.2.5). Find a differential equation for the quantity \(Q(t)\) of salt in the tank at time \(t\) prior to the time when the tank overflows and find the concentration \(K(t)\) (g/liter ) of salt in the tank at any such time.

Q=300-260e^-t/150 — Figure 4.2.4. \(Q=300-260e^{-t/150}\)

Solution We first determine the amount \(W(t)\) of solution in the tank at any time \(t\) prior to overflow. Since \(W(0) = 200\) and we’re adding 4 liters/min while removing only 2 liters/min, there’s a net gain of 2 liters/min in the tank; therefore,

\[ W(t) = 2t+200. \]

Since \(W(150)=500\) liters (capacity of the tank), this formula is valid for \(0 \le t \le 150\).

Now let \(Q(t)\) be the number of grams of salt in the tank at time \(t\), where \(0 \le t \le 150\). As in Example 4.2.3,

\begin{equation} Q' = \mbox{rate in}-\mbox{rate out}. \tag{4.2.7}\end{equation}

The rate in is

\begin{equation} \left({1\over 4}\ \mbox{g/liter}\,\right) \times (4\ \mbox{liters/min}\,) = 1\ \mbox{g/min}. \tag{4.2.8}\end{equation}

To determine the rate out, we observe that since the mixture is being removed from the tank at the constant rate of 2 liters/min and there are \(2t+200\) liters in the tank at time \(t\), the fraction of the mixture being removed per minute at time \(t\) is

\[ {2\over 2t+200} = {1\over t+100}. \]

We’re removing this same fraction of the salt per minute. Therefore, since there are \(Q(t)\) grams of salt in the tank at time \(t\),

\begin{equation} \mbox{rate out} = {Q(t)\over t+100}. \tag{4.2.9}\end{equation}

Alternatively, we can arrive at this conclusion by arguing that

\[ \begin{array}{lcl} \mbox{rate out} & = & (\mbox{concentration})\times(\mbox{rate of flow out}) =(\mbox{g/liter})\times(\mbox{liters/min})\\[10pt] &=&\dst{Q(t)\over2t+200}\times 2=\dst{Q(t)\over t+100}. \end{array} \]

Substituting (4.2.8) and (4.2.9) into (4.2.7) yields

\begin{equation} Q'=1-{Q\over t+100},\quad\text{ \rm so }\quad Q'+{1\over t+100} Q=1. \tag{4.2.10}\end{equation}

By separation of variables, \(1/(t+100)\) is a solution of the complementary equation, so the solutions of (4.2.10) are of the form

\[ Q={u\over t+100},\mbox{\quad where \quad}{u'\over t+100}=1,\mbox{\quad so \quad} u'=t+100. \]

Hence,

\begin{equation} u = {(t+100)^2\over 2}+c. \tag{4.2.11}\end{equation}

Since \(Q(0)=10\) and \(u=(t+100)Q\), (4.2.11) implies that

\[ (100)(10) = {(100)^2\over 2}+c, \]

\[ c=100(10)-{(100)^2\over 2} =-4000 \]

and therefore

\[ u = {(t+100)^2\over 2} -4000. \]

Hence,

\[ Q = {u\over t+200}= {t+100\over 2}-{4000\over t+100}. \]

Now let \(K(t)\) be the concentration of salt at time \(t\). Then

\[ K(t) = {1\over 4}-{2000\over(t+100)^2} \]

(Figure 4.2.6).

K(t) = 14-2000(t+100)^2 — Figure 4.2.6. \(K(t) = \dst{{1\over 4}-{2000\over(t+100)^2}}\)

4.2 Exercises

A thermometer is moved from a room where the temperature is \(70^\circ\)F to a freezer where the temperature is \(12^\circ F\). After 30 seconds the thermometer reads \(40^\circ\)F. What does it read after 2 minutes?

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\(\approx15.15^\circ\)F
A fluid initially at \(100^\circ\)C is placed outside on a day when the temperature is \(-10^\circ\)C, and the temperature of the fluid drops \(20^\circ\)C in one minute. Find the temperature \(T(t)\) of the fluid for \(t > 0\).

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\(\dst {T= -10+110 e^{-t \ln {11\over 9}}}\)
At 12:00 pm a thermometer reading \(10^\circ\)F is placed in a room where the temperature is \(70^\circ\)F. It reads \(56^\circ\) when it’s placed outside, where the temperature is \(5^\circ\)F, at 12:03. What does it read at 12:05 pm?

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\(\approx 24.33^\circ\)F
A thermometer initially reading \(212^\circ\)F is placed in a room where the temperature is \(70^\circ\)F. After 2 minutes the thermometer reads \(125^\circ\)F.
1. What does the thermometer read after 4 minutes?
2. When will the thermometer read \(72^\circ\)F?
3. When will the thermometer read \(69^\circ\)F?
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(a) \(91.30^\circ\)F (b) 8.99 minutes after being placed outside (c) never
An object with initial temperature \(150^\circ\)C is placed outside, where the temperature is \(35^\circ\)C. Its temperatures at 12:15 and 12:20 are \(120^\circ\)C and \(90^\circ\)C, respectively.
1. At what time was the object placed outside?
2. When will its temperature be \(40^\circ\)C?
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(a) 12:11:32 (b) 12:47:33
An object is placed in a room where the temperature is \(20^\circ\)C. The temperature of the object drops by \(5^\circ\)C in 4 minutes and by \(7^\circ\)C in 8 minutes. What was the temperature of the object when it was initially placed in the room?

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\(\dst {(85/3)^\circ C}\)
A cup of boiling water is placed outside at 1:00 pm. One minute later the temperature of the water is \(152^\circ\)F. After another minute its temperature is \(112^\circ\)F. What is the outside temperature?

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\(32^\circ\)F
A tank initially contains 40 gallons of pure water. A solution with 1 gram of salt per gallon of water is added to the tank at 3 gal/min, and the resulting solution dranes out at the same rate. Find the quantity \(Q(t)\) of salt in the tank at time \(t > 0\).

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\(\dst {Q(t) = 40(1-e^{-3 t/40})}\)
A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per gallon is added to the tank at 6 gal/min, and the resulting solution leaves at the same rate. Find the quantity \(Q(t)\) of salt in the tank at time \(t > 0\).

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\(\dst {Q(t) =30-20 e^{-t/10}}\)
A tank initially contains 100 liters of a salt solution with a concentration of .1 g/liter. A solution with a salt concentration of .3 g/liter is added to the tank at 5 liters/min, and the resulting mixture is drained out at the same rate. Find the concentration \(K(t)\) of salt in the tank as a function of \(t\).

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\(\dst {K(t)=.3-.2e^{-t/20}}\)
A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is drained out at 2 gal/min. Find the quantity of salt in the tank as it’s about to overflow.

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\(Q(50)=47.5\) (pounds)
Suppose water is added to a tank at 10 gal/min, but leaks out at the rate of 1/5 gal/min for each gallon in the tank. What is the smallest capacity the tank can have if the process is to continue indefinitely?

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50 gallons
A chemical reaction in a laboratory with volume \(V\) (in ft\(^3\)) produces \(q_1\) ft\(^3\)/min of a noxious gas as a byproduct. The gas is dangerous at concentrations greater than \(\overline c\), but harmless at concentrations \(\le \overline c\). Intake fans at one end of the laboratory pull in fresh air at the rate of \(q_2\) ft\(^3\)/min and exhaust fans at the other end exhaust the mixture of gas and air from the laboratory at the same rate. Assuming that the gas is always uniformly distributed in the room and its initial concentration \(c_0\) is at a safe level, find the smallest value of \(q_2\) required to maintain safe conditions in the laboratory for all time.

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\(\min q_2 = q_1/\overline{c}\)
A 1200-gallon tank initially contains 40 pounds of salt dissolved in 600 gallons of water. Starting at \(t_0=0\), water that contains 1/2 pound of salt per gallon is added to the tank at the rate of 6 gal/min and the resulting mixture is drained from the tank at 4 gal/min. Find the quantity \(Q(t)\) of salt in the tank at any time \(t > 0\) prior to overflow.

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\(\dst {Q = t+300-{234 \times 10^5\over(t+300)^2}, \quad 0 \le t \le 300}\)
Tank \(T_1\) initially contain 50 gallons of pure water. Starting at \(t_0=0\), water that contains 1 pound of salt per gallon is poured into \(T_1\) at the rate of 2 gal/min. The mixture is drained from \(T_1\) at the same rate into a second tank \(T_2\), which initially contains 50 gallons of pure water. Also starting at \(t_0=0\), a mixture from another source that contains 2 pounds of salt per gallon is poured into \(T_2\) at the rate of 2 gal/min. The mixture is drained from \(T_2\) at the rate of 4 gal/min.
1. Find a differential equation for the quantity \(Q(t)\) of salt in tank \(T_2\) at time \(t > 0\).
2. Solve the equation derived in (a) to determine \(Q(t)\).
3. Find \(\lim_{t\to\infty}Q(t)\).
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(a) \(\dst{Q'+{2\over25}Q=6-2e^{-t/25}}\) (b) \(Q=75-50e^{-t/25}-25e^{-2t/25}\) (c) 75
Suppose an object with initial temperature \(T_0\) is placed in a sealed container, which is in turn placed in a medium with temperature \(T_m\). Let the initial temperature of the container be \(S_0\). Assume that the temperature of the object does not affect the temperature of the container, which in turn does not affect the temperature of the medium. (These assumptions are reasonable, for example, if the object is a cup of coffee, the container is a house, and the medium is the atmosphere.)
1. Assuming that the container and the medium have distinct temperature decay constants \(k\) and \(k_m\) respectively, use Newton’s law of cooling to find the temperatures \(S(t)\) and \(T(t)\) of the container and object at time \(t\).
2. Assuming that the container and the medium have the same temperature decay constant \(k\), use Newton’s law of cooling to find the temperatures \(S(t)\) and \(T(t)\) of the container and object at time \(t\).
3. Find \(\lim._{t\to\infty}S(t)\) and \(\lim_{t\to\infty}T(t)\) .
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(a) \(T=T_m+(T_0-T_m)e^{-kt}+\dst{k(S_0-T_m)\over (k-k_m)}\left(e^{-k_mt}-e^{-kt}\right)\)

(b) \(T=T_m+k(S_0-T_m)te^{-kt}+(T_0-T_m)e^{-kt}\) (c) \(\lim_{t\to\infty}T(t)=\lim_{t\to\infty}S(t)=T_m\)
In our previous examples and exercises concerning Newton’s law of cooling we assumed that the temperature of the medium remains constant. This model is adequate if the heat lost or gained by the object is insignificant compared to the heat required to cause an appreciable change in the temperature of the medium. If this isn’t so, we must use a model that accounts for the heat exchanged between the object and the medium. Let \(T=T(t)\) and \(T_m=T_m(t)\) be the temperatures of the object and the medium, respectively, and let \(T_0\) and \(T_{m0}\) be their initial values. Again, we assume that \(T\) and \(T_m\) are related by Newton’s law of cooling,

\[ T'=-k(T-T_m). \tag*{\rm (A)} \]

We also assume that the change in heat of the object as its temperature changes from \(T_0\) to \(T\) is \(a(T-T_0)\) and that the change in heat of the medium as its temperature changes from \(T_{m0}\) to \(T_m\) is \(a_m(T_m-T_{m0})\), where \(a\) and \(a_m\) are positive constants depending upon the masses and thermal properties of the object and medium, respectively. If we assume that the total heat of the system consisting of the object and the medium remains constant (that is, energy is conserved), then

\[ a(T-T_0)+a_m(T_m-T_{m0})=0. \tag*{\rm (B)} \]
1. Equation (A) involves two unknown functions \(T\) and \(T_m\). Use (A) and (B) to derive a differential equation involving only \(T\).
2. Find \(T(t)\) and \(T_m(t)\) for \(t>0\).
3. Find \(\lim_{t\to\infty}T(t)\) and \(\lim_{t\to\infty}T_m(t)\).
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(a) \(T'=\dst{-k\left(1+{a\over a_m}\right)T +k\left(T_{m0}+{a\over a_m}T_0\right)}\) (b) \(\dst{T={aT_0+a_mT_{m0}\over a+a_m}+{a_m(T_0-T_{m0})\over a+a_m}e^{-k(1+a/a_m)t}}\),

\(\dst{T_m={aT_0+a_mT_{m0}\over a+a_m}+{a(T_{m0}-T_0)\over a+a_m}e^{-k(1+a/a_m)t}}\); (c) \(\lim_{t\to\infty}T(t)= \lim_{t\to\infty}T_m(t) =\dst{aT_0+a_mT_{m0}\over a+a_m}\)
Control mechanisms allow fluid to flow into a tank at a rate proportional to the volume \(V\) of fluid in the tank, and to flow out at a rate proportional to \(V^2\). Suppose \(V(0)=V_0\) and the constants of proportionality are \(a\) and \(b\), respectively. Find \(V(t)\) for \(t>0\) and find \(\lim_{t\to\infty}V(t)\).

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\(V=\dst{a\over b}\dst{V_0\over V_0-\left(V_0-a/b \right)e^{-at}}\), \(\lim_{t\to\infty}V(t)=a/b\)
Identical tanks \(T_1\) and \(T_2\) initially contain \(W\) gallons each of pure water. Starting at \(t_0=0\), a salt solution with constant concentration \(c\) is pumped into \(T_1\) at \(r\) gal/min and drained from \(T_1\) into \(T_2\) at the same rate. The resulting mixture in \(T_2\) is also drained at the same rate. Find the concentrations \(c_1(t)\) and \(c_2(t)\) in tanks \(T_1\) and \(T_2\) for \(t>0\).

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\(c_1=c\left(1-e^{-rt/W}\right)\), \(c_2=c\dst{\left(1-e^{-rt/W}-{r\over W}te^{-rt/W}\right)}\).
An infinite sequence of identical tanks \(T_1\), \(T_2\), …, \(T_n\), …, initially contain \(W\) gallons each of pure water. They are hooked together so that fluid drains from \(T_n\) into \(T_{n+1}\,(n=1,2,\cdots)\). A salt solution is circulated through the tanks so that it enters and leaves each tank at the constant rate of \(r\) gal/min. The solution has a concentration of \(c\) pounds of salt per gallon when it enters \(T_1\).
1. Find the concentration \(c_n(t)\) in tank \(T_n\) for \(t>0\).
2. Find \(\lim_{t\to\infty}c_n(t)\) for each \(n\).
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(a) \(\dst{c_n=c\left(1-e^{-rt/W}\sum_{j=0}^{n-1}{1\over j!}\left(rt\over W\right)^j\right)}\) (b) \(c\) (c) \(0\)
Tanks \(T_1\) and \(T_2\) have capacities \(W_1\) and \(W_2\) liters, respectively. Initially they are both full of dye solutions with concentrations \(c_{1}\) and \(c_2\) grams per liter. Starting at \(t_0=0\), the solution from \(T_1\) is pumped into \(T_2\) at a rate of \(r\) liters per minute, and the solution from \(T_2\) is pumped into \(T_1\) at the same rate.
1. Find the concentrations \(c_1(t)\) and \(c_2(t)\) of the dye in \(T_1\) and \(T_2\) for \(t>0\).
2. Find \(\lim_{t\to\infty}c_1(t)\) and \(\lim_{t\to\infty}c_2(t)\).
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Let \(c_\infty=\dst{c_1W_1+c_2W_2\over W_1+W_2}\), \(\alpha=\dst{c_2W_2^2-c_1W_1^2\over W_1+W_2}\), and \(\beta=\dst{W_1+W_2\over W_1W_2}\). Then:

(a) \(c_1(t)=c_\infty+\dst{\alpha\over W_1}e^{-r\beta t}\), \(c_2(t)=c_\infty-\dst{\alpha\over W_2}e^{-r\beta t}\)

(b) \(\lim_{t\to\infty}c_1(t)=\lim_{t\to\infty}c_2(t)=c_\infty\)
L Consider the mixing problem of Example 4.2.3, but without the assumption that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Assume instead that the distribution approaches uniformity as \(t\to\infty\). In this case the differential equation for \(Q\) is of the form

\[ Q'+{a(t)\over150}Q=2 \]

where \(\lim_{t\to\infty}a(t)=1\).
1. Assuming that \(Q(0)=Q_0\), can you guess the value of \(\lim_{t\to\infty}Q(t)\)?.
2. Use numerical methods to confirm your guess in the these cases:
  
  \[ \mbox{\part{i}\; } a(t)=t/(1+t) \mbox{\quad \part{ii}\; } a(t)=1-e^{-t^2} \mbox{\quad \part{iii}\; } a(t)=1-\sin(e^{-t}). \]
L Consider the mixing problem of Example 4.2.4 in a tank with infinite capacity, but without the assumption that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Assume instead that the distribution approaches uniformity as \(t\to\infty\). In this case the differential equation for \(Q\) is of the form

\[ Q'+{a(t)\over t+100}Q=1 \]

where \(\lim_{t\to\infty}a(t)=1\).
1. Let \(K(t)\) be the concentration of salt at time \(t\). Assuming that \(Q(0)=Q_0\), can you guess the value of \(\lim_{t\to\infty}K(t)\)?
2. Use numerical methods to confirm your guess in the these cases:
  
  \[ \mbox{\part{i}\; } a(t)=t/(1+t)\; \mbox{\; \part{ii}\; } a(t)=1-e^{-t^2} \mbox{\; \part{iii}\; } a(t)=1+\sin(e^{-t}). \]