Question

dT/dt = k(T − A), where T is the temperature of the object, t is time, k is the proportionality constant, and A is the constant ambient temperature. T (t) = A + Ce^kt is the general solution. Apply the solution to the following scenario: A Police Department officer discovered a corpse in a downtown alley at 1130pm on a night where the constant temperature was 40 degrees Fahrenheit. As she had been trained to do, she immediately recorded the body temperature of the corpse: 93 degrees Fahrenheit. She took the temperature of the corpse again at midnight: 90 degrees Fahrenheit. Assuming that the body’s temperature at the time of death was 98.6 degrees Fahrenheit, at what time did death occur? Give an exact answer and an estimate to the nearest minute.

Answer #1

Newton's law of cooling is: du/dt = -k (u-T) where u(t) is
temperature of an object, t is in hours, T is a constant ambient
temperature, and k is a positive constant.
Suppose a building loses heat in accordance with Newton's law of
cooling. Suppose that the rate constant k has the value 0.15 hr^-1
. Assume that the interior temperature is Ti = 77F, when the
heating system fails. If the external temperature is T = 5F, how
long...

1. Solve the given initial value problem. dy/dt = (t^3 +
t)/(y^2); y(0) = 2 .
2. We know from Newton’s Law of Cooling that the rate at which a
cold soda warms up is proportional to the difference between the
ambient temperature of the room and the temperature of the drink.
The differential equation corresponding to this situation is given
by y' = k(M − y) where k is a positive constant. The solution to
this equation is given...

Newton’s law of cooling states that dx/dt = −k(x − A) where x is
the temperature, t is time, A is the ambient temperature, and k
> 0 is a constant. Suppose that A = A0cos(ωt) for some constants
A0 and ω. That is, the ambient temperature oscillates (for example
night and day temperatures). a) Find the general solution. b) In
the long term, will the initial conditions make much of a
difference? Why or why not?

Question B:
Newton's law of cooling states
dθ/dt = −k (θ−T)
where ? is the temperature at time t, T is the constant
surrounding temperature and k is a constant.
If a mass with initial temperature, θ0, of 319.5 K is
placed in a surroundings of 330.5 K, and k is 0.011 s-1
, what is its temperature after 4.7 minutes? Give your answer to 4
significant figures and remember to use units.
____________

If the temperature difference DT between an object and
it surroundings is not too great, the rate of cooling or warming
obeys Newton’s Law of Cooling,
d(DT)/dt = – K DT
where K is a constant. (a) Why is there a
minus sign on the right-hand side of the equation? (b) On
what factors does K depend and what are its units?
(c) It DTo is the temperature difference
at time to, what is the temperature difference at
time t...

police arrive at a murder scene at 1:30 am and immediately
record the body's temperature which was 93 degrees Fahrenheit. at
2:00 am, after thoroughly inspecting and fingerprinting the area,
they again took the temperature of the body which had dropped to 83
degrees F. the temperature of the crime scene has remained at a
constant 70 degrees F. Determine when the person was murdered.
(assume that the victim was healthy at the time of death. that is,
assume that...

Newton’s Law of Cooling tells us that the time rate of chnge in
temperature T(t) of a body immersed in a medium of constant
temperature A is proportional to the difference A − T.The DE
modeling this is dT dt = k(A − T). A cup of hot chocolate is
initially 170◦ F and is left in a room with an ambient temperature
of 70◦ F. Suppose that at time t = 0 it is cooling at a rate of...

Newton's Law of Cooling tells us that the rate of change of the
temperature of an object is proportional to the temperature
difference between the object and its surroundings. This can be
modeled by the differential equation dTdt=k(T−A)dTdt=k(T-A), where
TT is the temperature of the object after tt units of time have
passed, AA is the ambient temperature of the object's surroundings,
and kk is a constant of proportionality.
Suppose that a cup of coffee begins at 179179 degrees and,...

The Arrhenius equation shows the relationship between the rate
constant k and the temperature T in kelvins and is typically
written as k=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K),
A is a constant called the frequency factor, and Ea is the
activation energy for the reaction. However, a more practical form
of this equation is lnk2k1=EaR(1T1−1T2) which is mathmatically
equivalent to lnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate
constants for a single reaction at two different absolute...

Using Exponentials and Logarithms, show that
T2/T1
=(V1/V2)k-1
where p is pressure T is temperature and V is volume k is
is a constant.

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