Question

If the temperature difference D*T* between an object and
it surroundings is not too great, the rate of cooling or warming
obeys *Newton’s Law of Cooling*,

*d(*D*T)/dt = – K* D*T*

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 D*T**o* is the temperature difference
at time *t**o*, what is the temperature difference at
time *t* later? (d) A medical examiner is called to a crime
scene where a murder has occurred. Upon arriving at 1 AM the
examiner measures the body temperature to be 79o F. An hour later
the examiner measures the body temperature to be 77.5o F. A
detective tells the examiner that the read-out on the computer
controlled thermostat in the room shows that the temperature in the
room has remained constant at a 70o F for the past two days. The
detective asks the examiner for the time of death. What time does
the examiner tell the detective?

Answer #1

A detective is called to the scene of a crime where a dead body
has just been found. She arrives on the scene at 10:23PM and begins
her investigation. Immediately, the temperature of the body is
taken and found to be 80℉. The detective checks the thermostat and
finds that the room has been kept at a constant 68℉. After the
evidence from the crime scene is collected, the temperature of the
body is taken once more exactly one hour...

The coroner arrives at a murder scene at 9:00 pm. He immediately
determines that the temperature of the body is 83◦F. He waits one
hour and takes the temperature of the body again; it is 81◦F. The
room temperature is 68◦F. When was the murder
committed? Assume the man died with a body temperature
98◦F. (Hint: Newton’s Law of Cooling)

Cooling:
The body of an apparent victim of a crime is discovered
by detectives at 9 AM, at which time the body temperature was
measured to be 88°. Two hours later the temperature was 82°. Using
Newton’s Law of Cooling, determine the time of death if the room
temperature was a constant 75°.

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?

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 states that the rate of change of the
temperature T of an object is proportional to the temperature
difference between the temperature S of the surroundings and the
temperature T. dT dt = k(S − T) A cup of tea is prepared from
boiling water at 100 degrees and cools to 60 degrees in 2 minutes.
The temperature in the room is 20 degrees. 1. What will the
temperature be after 15 minutes?

15. Newton’s Law of Cooling. Newton’s law of cooling
states that the rate of change in the temperature T(t) of
a body is proportional to the difference between the
temperature
of the medium M(t) and the temperature of the
body. That is,
dT/dt = K[M(t) - T(t)] ,
where K is a constant. Let K = 0.04 (min)-1 and the
temperature
of the medium be constant, M(t) = 293 kelvins.
If the body is initially at 360 kelvins, use Euler’s...

Newton's law of cooling states that the temperature of an object
changes at a rate proportional to the different between its
temperature and that of its surroundings. Suppose that the
temperature of a cup of coffee obeys Newton's law of cooling. If
the coffee has a temperature of 200 degrees F when freshly poured,
and 1 min later has cooled to 190 degrees F in a room at 70 degrees
F, determine when the coffee reaches a temperature of 150...

This question is about Newton’s law of cooling, which states
that the temperature of a hot object decreases proportionally to
the difference between its temperature and the temperature of the
surroundings. This can be written as dT dt = −k(T − Ts), where T is
the temperature, t is time, k is a constant and Ts is the
temperature of the surroundings. For this question we will assume
that the surroundings are at a constant 20◦ and A that the...

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.
____________

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