Asked by satan

Newton’s law of cooling states that for a cooling substance with initial temperature T0
T
0
, the temperature T(t)
T
(
t
)
after t minutes can be modeled by the equation T(t)=Ts+(T0−Ts)e−kt
T
(
t
)
=
T
s
+
(
T
0

T
s
)
e

k
t
, where Ts
T
s
is the surrounding temperature and k is the substance’s cooling rate.

A liquid substance is heated to 80°C
80
°
C
. Upon removing it from the heat it cools to 60°C
60
°
C
in 15 minutes.

What is the substance’s cooling rate when the surrounding air temperature is 50°C
50
°
C
?

Round the answer to four decimal places.



0.0687 <my choice

0.0732

0.0813

0.0872

Answers

Answered by MathMate
Please show your work to find where your calculations went wrong (if applicable).
Answered by Michelle Hicks
Newton’s law of cooling states that for a cooling substance with initial temperature T0 , the temperature T(t) after t minutes can be modeled by the equation T(t)=Ts+(T0−Ts)e−kt , where Ts is the surrounding temperature and k is the substance’s cooling rate.

A liquid substance is heated to 80°C . Upon being removed from the heat, it cools to 60°C in 12 min.

What is the substance’s cooling rate when the surrounding air temperature is 50°C ?

The substances cooling rate when the surrounding air temperature is 50C is 0.0916.

0.0687
0.0732
0.0813
0.0916 <------- Correct answer!!
Answered by Thomas
0.0916 Would be the correct answer to this Question.

I have taken the test and passed with a 100%

Hope this helps.
Answered by lynn
if you see the original post says 15 mins not 12 so 0.0916 is not the correct answer or even an option
Answered by Bob
Super man

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