Question

In 1701, Issac Newton proved his Law of Cooling: T(t)
=Ae^{kt} +T_{a}, which is an exponential model that
relates the temperature of an object *T* as a function of
time *t* (we will use minutes) that is placed in an
environment with ambient temperature T_{a}.

Suppose a cup of hot coffee is served at 160◦F and placed in a room with an ambient temperature 75◦. After 5 minutes, the cup of coffee has a temperature of 131◦F.

a) Create a model using the Newton’s Law of Cooling template.

b) What will be the temperature of the coffee after 12 minutes?

c) How long until the coffee reaches 77◦F?

Answer #1

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...

According to Newton's Law of Cooling
A cup of coffee with temperature of 130F is placed in a freezer
with temperature 0F. After 5 minutes, the temperature of the coffee
is 87F. Find the coffee's temperature
after 10 minutes.

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?

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...

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,...

(1 point) Newton's Law of Cooling states that the rate of
cooling of an object is proportional to the temperature difference
between the object and its surroundings. Suppose t is time, T is
the temperature of the object, and Ts is the surrounding
temperature. The following differential equation describes Newton's
Law dT/dt=k(T−Ts), where k is a constant. Suppose that we consider
a 95∘C cup of coffee in a 25∘C room. Suppose it is known that the
coffee cools at a...

Newton’s Law of Cooling and the Ornstein-Uhlenbeck
Process
The Law of Cooling says the temperature difference between an
object (say a hot cup of coffee) and the ambient temperature (the
temperature in the room) declines exponentially:
If T(t) is the temperature of the object at time t, we have the
ODE:
d/dt(T(t) – Troom) = b (T(t) –
Troom) b < 0, Troom is a
constant.
Equivalently, dT/dt = b (T – Troom)
T(0) is the starting temperature of the...

A student brings a cup of hot (190° ) coffee to her evening
Differential Equations course. Unfortunately, the classroom is 80°
that day because the air conditioner is broken. Ms. Jensen opens
the door and the classroom begins to cool at a constant rate of 10°
per hour. The student is so absorbed in the class that she forgets
about her coffee and it sits on her desk for 50 minutes as the
classroom cools down. In this problem the...

Suppose a cup of coffee is 160 degrees right after the coffee is
poured. You are only comfortable drinking coffee when it cools to
120 degrees. You check after five minutes, and the coffee is still
hot (135 degrees!). When will you be able to first comfortably sip
the coffee? Note: The equation for Newton’s law of cooling is ?? =
? (? − ??); its solution is:T(t) = Ts + (T0 – Ts) e^kt. Assume the
surrounding temperature is...

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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