Newton’s law of cooling states that for a cooling substance with initial temperature T0

Question

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

Answer 1

Please show your work to find where your calculations went wrong (if applicable).

Answer 2

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!!

Answer 3

0.0916 Would be the correct answer to this Question.

I have taken the test and passed with a 100%

Hope this helps.

Answer 4

if you see the original post says 15 mins not 12 so 0.0916 is not the correct answer or even an option

Answer 5

Super man