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
Please show your work to find where your calculations went wrong (if applicable).
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!!
Super man
To find the substance's cooling rate, we can use the equation for Newton's law of cooling: T(t) = Ts + (T0 - Ts)e^(-kt).
Given information:
- Initial temperature, T0 = 80°C
- Temperature after t minutes, T(t) = 60°C
- Time, t = 15 minutes
- Surrounding temperature, Ts = 50°C
Substituting the values into the equation, we can solve for the cooling rate, k.
60 = 50 + (80 - 50)e^(-15k)
Simplifying the equation, we have:
10 = 30e^(-15k)
Dividing both sides by 30:
1/3 = e^(-15k)
To find the natural logarithm of both sides:
ln(1/3) = -15k
Now, divide both sides by -15:
ln(1/3) / -15 = k
Using a calculator to evaluate the left-hand side:
k ≈ 0.0687 (rounded to four decimal places)
Therefore, the substance's cooling rate when the surrounding air temperature is 50°C is approximately 0.0687.
0.0916 Would be the correct answer to this Question.
I have taken the test and passed with a 100%
Hope this helps.