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Mathematics LibreTexts

4.2: Cooling and Mixing

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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 Tm(t), the rate of change of T at time t is proportional to T(t)Tm(t); thus, T satisfies a differential equation of the form

T=k(TTm).

Here k>0, since the temperature of the object must decrease if T>Tm, or increase if T<Tm. 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 Tm. 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 is 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 is a huge cauldron of molten metal. (For more on this see Exercise 4.2.17.)

To solve Equation ???, we rewrite it as

T+kT=kTm.

Since ekt is a solution of the complementary equation, the solutions of this equation are of the form T=uekt, where uekt=kTm, so u=kTmekt. Hence,

u=Tmekt+c,

so

T=uekt=Tm+cekt.

If T(0)=T0, setting t=0 here yields c=T0Tm, so

T=Tm+(T0Tm)ekt.

Note that TTm decays exponentially, with decay constant k.

Example 4.2.1

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

Solution

Here T0=400 and Tm=25, so Equation ??? becomes

T=25+375ekt.

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

200=25+375e4k;

hence,

e4k=175375=715.

Taking logarithms and solving for k yields

k=14ln715=14ln157.

Substituting this into Equation ??? yields

T=25+375et4ln157

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

T(8)=25+375e2ln157=25+375(715)2107C.

Example 4.2.2

An object with temperature 72F is placed outside, where the temperature is 20F. At 11:05 the temperature of the object is 60F and at 11:07 its temperature is 50F. 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 t0=0 of the time scale to be 11:05 so that T0=60. We must determine the time τ when T(τ)=72. Substituting T0=60 and Tm=20 into Equation ??? yields

T=20+(60(20))ekt

or

T=20+80ekt.

fig040201.svg
Figure 4.2.1 : T=25+375e(t/4)ln15/7

We obtain k from the stated condition that the temperature of the object is 50F 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 Equation ??? to obtain

50=20+80e2k

This is shown in Figure 4.2.2 .

Hence,

e2k=7080=78.

Taking logarithms and solving for k yields

k=12ln78=12ln87.

Substituting this into Equation ??? yields

T=20+80et2ln87,

and the condition T(τ)=72 implies that

72=20+80eτ2ln87;

hence,

eτ2ln87=9280=2320.

Taking logarithms and solving for τ yields

τ=2ln2320ln872.09 min.

fig040202.svg
Figure 4.2.2 : 20+80et2ln87

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

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 4.2.22 and 4.2.23 deal with situations where this isn’t so, but the distribution of salt becomes approximately uniform as t.

Example 4.2.3

A tank initially contains 40 pounds of salt dissolved in 600 gallons of water. Starting at t0=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 ).

  1. 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).
  2. Find limtQ(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

Q=rate inrate out.

fig040203.svg
Figure 4.2.3 : A mixing problem

The rate in is

(12 lb/gal)×(4 gal/min)=2 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 are 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

rate out=(concentration)×(rate of flow out)=(lb/gal)×(gal/min)=Q(t)600×4=Q(t)150.

We can now write Equation ??? as

Q=2Q150.

This first order equation can be rewritten as

Q+Q150=2.

Since et/150 is a solution of the complementary equation, the solutions of this equation are of the form Q=uet/150, where uet/150=2, so u=2et/150. Hence,

u=300et/150+c,

fig040204.svg
Figure 4.2.4 : Q=300260et/150

so

Q=uet/150=300+cet/150

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

Q=300260et/150.

Solution b

From Equation ???, we see that that limtQ(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 t0=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.

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 are 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 0t150.

Now let Q(t) be the number of grams of salt in the tank at time t, where 0t150. As in Example 4.2.3

Q=rate inrate out.

fig040205.svg
Figure 4.2.5 : Another mixing problem

The rate in is

(14 g/liter)×(4 liters/min)=1 g/min.

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

22t+200=1t+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,

rate out=Q(t)t+100.

Alternatively, we can arrive at this conclusion by arguing that

rate out=(concentration)×(rate of flow out)=(g/liter)×(liters/min)=Q(t)2t+200×2=Q(t)t+100.

Substituting Equation ??? and Equation ??? into Equation ??? yields

Q=1Qt+100,soQ+1t+100Q=1.

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

Q=ut+100,whereut+100=1,sou=t+100.

Hence,

u=(t+100)22+c.

Since Q(0)=10 and u=(t+100)Q, Equation ??? implies that

(100)(10)=(100)22+c,

so

c=100(10)(100)22=4000

and therefore

u=(t+100)224000.

Hence,

Q=ut+200=t+10024000t+100.

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

K(t)=142000(t+100)2

This is shown in Figure 4.2.6 .

fig040206.svg
Figure 4.2.6 : K(t)=142000(t+100)2

This page titled 4.2: Cooling and Mixing is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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