Question

A mug of coffee, with a temperature ?℃ is made and left to cool in a room with a temperature of 25℃. The rate at which the coffee cools is proportional to the difference in temperature between the coffee and the room. Initially the coffee is at a temperature 85℃. 10 minutes later the coffee is at 55℃. Determine the temperature, to 1 decimal place, of the coffee after 15 minutes.

Answer #1

A covered mug of coffee originally at 190 degrees Fahrenheit, if
left for t hours in a room whose temperature is 60
degrees, will cool to a temperature of
60 +
130e−1.7t
degrees. Find the temperature of the coffee after the following
amounts of time. (Round your answers to the nearest degree.)
(a) 15 minutes
°F
(b) half an hour
°F

A cup of coffee has a temperature of 200F. It is placed in a
room that has temperature 70 F. After 15 minutes, the temperature
of the coffee is 150F.
a) Model the temperature of the cup of coffee at time t.
b) How long will it take for the coffee to cool down to 100◦
F?

Water is boiled and is put to cool. The temperature of the air
in the room is rising linearly in relation to the function
T(t)=30+0.01t, where t is minutes and T is Celsius. Suppose this
satisfies Newton's Law of Cooling: the change of temperature of the
water is proportional to the difference of the water temperature
and the room’s ambient temperature. We take the water's temperature
10 minutes after and find that it's 81*C. Graph and explain T.
Establish and...

A mug of beer chilled to 30 degrees, if left in a 72-degree
room, will warm to a temperature of
T(t) = 72 −
42e−3.5t
degrees in t hours.
(a) Find T(0.45) and
T'(0.45).
(Round your answers to one decimal place.)
T(0.45)
=
°
T'(0.45)
=
°
Interpret your answers.
After 27 minutes, the temperature of the beer
is degrees and is increasing at the rate
of degrees per hour.
(b) Find T(1) and
T'(1).
(Round your answers to one decimal place.)...

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?

A 0.217-kg coffee mug is made from a material that has a
specific heat capacity of 989 J/kg C° and contains 0.315 kg of
water. The cup and water are at 25.8 °C. To make a cup of coffee, a
small electric heater is immersed in the water and brings it to a
boil in two minutes. Assume that the cup and water always have the
same temperature and determine the minimum power rating of this
heater.

You want to cool 0.2 kg of coffee, initially at temperature Th =
80° C, with ice initially at Tc = 0° C. The specific heat of ice is
about 2108 J/kg K, and its latent heat of melting is about 334, 000
J/kg. You may take the specific heats of liquid water and coffee to
be the same: 4187 J/kg K.
A) Assume the coffee and ice form a closed system. You want them to
equilibrate at 40° C....

(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 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,...

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...

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