Differential Calculus

 

1. 

The problem statement, all variables and given/known data 

a) An object at 200 degrees F is put in a room at 60 degrees F.The temperature of  the room decreases at the constant rate of 1 degree every 10 minutes. The body cools to 120 degrees F in 30 minutes. How long will it take for the body to cool to 90 degrees F? b) Show that the solution of the pertinent initial value problem which models the situation is: T(t) = 60 + 140e^(kt) + [(e^(kt) - kt - 1)/(10k)] c) Set-up an equation from which you can solve for k. d) Set-up an equation from which the required cooling time can be found. 2.

Relevant equations 

Newton's Law of Cooling: T'(t) = K(T(t) - T0)

Note: T is in minutes 3. The attempt at a solution 

a) This is variable seperable dT/dt = K(T(t) - T0) œdT/(T(t) - T0) = œk dt + C ln (T(t) - T0) = kt + C (T(t) - T0) = ce^(kt) T(t) = ce^(kt) + T0  At T(0) = 200, and T0 = 60 200 = ce^(K*0) + 60 c = 140 T(t) = 140e^(kt) + 60 This is where I get stuck. I'm not really sure where to go next. I'm mainly confused by the fact that room temperature is decreasing as well.Yes, that's the problem. The differential equation you started with, dT/dt = K(T(t) - T0), assumes that the

 

 

ambient temperature, T0, remains constant. The function that represents the ambient temperature is Ta = -t/10 + 60. You need to work that into the differential equation instead of T0.

A cup of coffee is poured from a potwhose contents are into a non-insulated cup in a room at . After a minute, the coffee has cooled to . How much time is required before the coffee reaches a drinkable temperature of ? 1.  Enter the equation.

2.  > eq1:= T = exp(k*t+C) +20;

3.  Solve for C  by using the initial temperature.

4. 

> eq2:= subs({t=0,T=95},eq1);

> C:=evalf(solve(eq2,C));

5.  Solve for k  by using the temperature at t =1. =1.

6. 

> eq3:=subs({t=1,T=90},eq1);

>k:=solve(eq3,k);

7.  Look at the original equation and notice that the only variable is t . Proceed to enter desired temperature and solve for t .

8. 

> eq1;

> eq4:=subs(T=65,eq1);

>solve(eq4,t);

 

 

The

time it takes for the coffee to reach

is 7.4 minutes.

Newton's Law of Cooling From experimental observations it is known that (up to a ``satisfactory'' approximation) the surface temperature of an object changes at a rate proportional to its relative temperature. That is, the difference between its temperature and the temperature of the surrounding surr ounding environment. This is what is known as Newton's law of cooling. Thus, if

is the temperature of the object at time t , then we have

where S  is the temperature of the surrounding environment. A qualitative study of this  phenomena will show that k  >0. This is a first order linear differential equation. The solution, under the initial condition

, is given by

Hence,

, which implies

equation makes it possible to find k  if the interval of time vice-versa.

This

is known and

Time of Death Suppose that a corpse was discovered in a motel room at midnight and its temperature was . The temperature of the t he room is kept constant at . Two hours later the temperature of the corpse dropped to . Find the time of death. Example:

 

 

Solution:

First we use the observed temperatures of the corpse to find the constant k .

We have

. In order to find the time of death we need to remember that the temperature of a corpse at time of death is (assuming the dead person was not sick!). Then we have

which means that the death happened around a round 7:26 P.M. On

a hot day a thermometer is takend outside from an air conditioned room where the temperature is 21C. After one minute it reads 27 and after two it reads 30. What is the temperature outside. The basic scheme is your standard 'growth and decay' setup. And you are right that: dT/dt = k(T - C), where, I assume, T is the reading on the thermometer, and C is the outside temperature, which we don't know. What you have there is a Differential Equation -- an equation involving one or more derivatives of an unknown function. In this case, the unknown function function is T(t) and we solve solve the equation by Separation of Variables. The variables are T and t. (Not T and C, even though you are supposed to find C. For the purposes of this equation, C is a constant.) After separating, the equation looks like this: dT ----- = k dt T-C and we integrate each side. ln(T - C) = kt + A

(A constant of integration.)

Now exponentiate to get: T - C = e^(kt + A) and rewrite based on exponent properties: T - C = e^kte^A T - C = a e^kt