Lemigio K.
asked • 12/22/21DIFFERENTIAL EQUATIONS
A hot cup of coffee at 90°C is placed on a table in the kitchen where the room temperature is 20°C. Using Newton's law of cooling, find at what rate the temperature of the coffee was decreasing when it was placed on the table. Give correct units.
Additional information: After 10 minutes the temperature of the coffee was 60°C.
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2 Answers By Expert Tutors
Luke J. answered • 12/23/21
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Experienced High School through College STEM Tutor
Given:
T(0) = 90°C
Ta = 20°C
T(10) = 60°C
Newton's Law of Cooling: dT/dt = -k * ( T(t) - Ta )
Find:
dT/dt ( t = 0 ) = ? °C / min
Solution:
T(t) = Ta + ( T(0) - Ta ) * e-kt. T(t) = 20 + 70 * e-kt
T(10) = 60 = 20 + 70 * e-10k 4 = 7 * exp(-10k)
4 * exp( +10k) = 7 [I don't like negatives if they don't appear to help later so I multipled both sides of the previous equation by e10k, as a side note, e^something and exp(something) are the same function, just easier to type since I'm doing this on my phone currently and don't have a good computer close by Lol]
exp( 10k ) = 7/4 = 1.75 10 k = ln( 1.75 )
k = 0.1 * ln( 1.75 ) ∴ k ≈ 0.0560 1/min
To the original Newton's Law of Cooling:
dT/dt = -k * ( T(t) - Ta )
dT/dt ( t = 0 ) = - 0.0560 * ( 90 - 20 ) °C/min
∴ dT/dt ( t = 0 ) = - 3.92 °C/min
Thus, at the initial setting on the table, the coffee was losing about 3.92°C of temperature every minute due to Newton's Law of Cooling and would slowly lose less and less temperature because it would be forever approaching the ambient room temperature
It would eventually reach a loss of approximately 0°C every minute because it would be so close to the room temperature, the temperature difference would be miniscule, numerical in size, but approaching zero.
This holds so long as it reminds undisturbed and if Newton's Law of Cooling were a 100% effective model of reality (of which it is sadly not because the system, in reality, would not be isolated from the other modes of heat transfer that could be present in the room unless you had some REALLY high tech and insulated equipment)
I hope this helps! Message me in the comments with any questions, comments, or concerns you may have!
Lemigio K.
Thank you so much Luke. This is the same answer I already have. I was however not convinced, hence looked for second opinion. You really helped with your explanation. I was of the notion that it will lose temperature at a constant rate. Hence, plugging in 3.9°C/min into some of the already given information was not making sense. Kudos once again. Your explanation is 100% superb.
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12/23/21
Lemigio K.
On another note Luke. I find out that if I model my DE as: dT/dt = k[T(t) - T(s)] not as dT/dt = -k(T(t) - T(s) I will still arrive at the same answer. So do I necessarily have to have the "-" sign in front of k?
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12/23/21
Luke J.
No problem! I'm glad I could help, reassure you of your answer, and teach you something new! If it were a constant, at all times loss of 3.9°C/min, the T(t) would be a linear line with a negative slope; however, plotting T(t) = 20 + 70 * exp( - 0.0560 * t), you'll see the slope changing and nearly zeroing out and the graph "flattening out" per say as time increases towards +infinity, I just gave the word description of that in my little soapbox at the end of my answer Lol
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12/23/21
Lemigio K.
Wonderful explanation Luke. I love this. The layout of your work is superb. It resonates with who I am as a person. :)
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12/23/21
Luke J.
Hmmmmmm, how did you arrive at the same answer? I agree you arrive at the same numerical answer, yet with the wrong sign convention. It defeats the purpose of Newton's Law of Cooling, the system needs to LOSE temperature, not continuously compound gain temperature. Essentially swapping the negative sign, you'd be evaluating the temperature "GAINED" by the "surroundings"; not having - on the k changes the perspective of what you are observing. I quote what I do because in the set up of this problem, the surrounding should NOT gain temperature, it's assumed to be too massive for the ambient fluid temperature to be altered by the cooling of the coffee
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12/23/21
Lemigio K.
Lol, if we don't put the "-" sign in front of the k,then you have T = 20 + 70e^(kt) and k = -0.05... Now if we put "-" sign, we have T = 20 + 70e^(-kt) and k = 0.05.. So having them both as I stated above you will arrive at the same numerical figure for any given T(t) = ? However, what you are saying makes a lot of sense. Hence, I will stick to having the "-" sign to depict that we are dealing with a cooling object not an object that is gaining heat from its surroundings.
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12/23/21
Yefim S. answered • 12/22/21
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Math Tutor with Experience
T(t) = Ts + (To – Ts) e-kt; T(10) = 60 = 20 + (90 - 20)e-10k; 40 = 70e-10k; k = - 0.1ln(4/7) = 0.056 min-1
Lemigio K.
Isn't -0.0556 the value for k? Now T = 20 + 70e^-kt. Don't we take the derivative of the general solution and plug in 0 for t since we are looking for the rate of change of temperature w.r.t. time? I already have the answer as 3.9°C. I am looking for second opinion.
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12/23/21
Yefim S.
k = 0.056 1/min
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12/23/21
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Luke J.
There is one missing piece of crucial information because Newton's Law of Cooling has a scaling constant multiplied onto the difference between the temperature of the cooling object and the ambient fluid temperature surrounding the cooling object. dT/dt = -k * ( T(t) - Ta ) This will allow you to get your answer but something needs told about that constant k because otherwise, there'll be WAY too many assumptions being made on how to get said k (one is typically told about a later temperature at a later time and can back calculate for the constant k)12/22/21