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Newton's Law of Cooling is based on the idea that the temperature of an object changes exponentially over time in the direction of the temperature of the surrounding medium.

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In equation form, the temperature of an object at any time t is given by $u(t)$.
$u(t) = T + (u_o - T) e^{kt}$
$k \lt 0$ constant
$T$ is the constant surrounding temperature
$u_o$ is the initial temperature of the object

Notes

1.In these equations it seems like we are assuming that the initial object temperature is greater than the surrounding temperature. However, this equation also holds when the initial object temperature is less than the surrounding temperature. In this case, the equation can be called Newton's Law of Warming.
2. You may also see the term surrounding medium or ambient temperature instead of surrounding temperature. This usually refers to air but it can be anything in general.
3. This is the same equation as exponential decay except shifted up T units.
4. There are many, many ways to write this equation. Another common way is $T(t) = T_s + (T_o - T_s) e^{kt}$, where $T_s$ is the surrounding temperature and $T_o$ is the initial temperature. You may also see the exponential written $e^{-kt}$ requiring $k \gt 0$. Check with your instructor and textbook to see what they require.

Before we try some practice problems, here is a great video with explanation and an example.

Michel vanBiezen - PreCalculus - Logarithmic & Exponential Functions (16 of 20) Newton's Law of Cooling [4min-55secs]

video by Michel vanBiezen

Practice

A thermometer reading 8^oC is brought into a room with a constant temperature of 35^oC. If the thermometer reads 15^oC after 3 minutes, determine the exponential equation and calculate what the thermometer will read after being in the room for 5 minutes.

Problem Statement

Final Answer

$\displaystyle{ u(t) = 35 - 27e^{t\ln(20/27)/3} }$ and $\displaystyle{ u(5) = 35 - 27e^{5\ln(20/27)/3} }$

Problem Statement

Solution

Given: $u_o = 8, T = 35$
To determine k, we will use $u(3) = 15$

$u(t) = T + (u_o - T) e^{kt}$

$u(t) = 35 + (8 - 35) e^{kt}$

$u(3) = 35 - 27 e^{k(3)} = 15$

$ - 27 e^{k(3)} = -20 $

$\displaystyle{ \frac{20}{27} = e^{3k} }$

$\displaystyle{ \ln\frac{20}{27} = \ln e^{3k} = 3k }$

$\displaystyle{ k = \frac{\ln(20/27)}{3} }$

$\displaystyle{ u(t) = 35 - 27e^{t\ln(20/27)/3} }$
To determine the temperature when t=5, we just plug 5 in for t in the equation for u(t) to get $\displaystyle{ u(5) = 35 - 27e^{5\ln(20/27)/3} }$
If you are allowed to use a calculator, the approximate answers are
$\displaystyle{ u(t) = 35 - 27e^{-0.100t} }$ and $ u(5) \approx 18.63^oC $

Final Answer
\(\displaystyle{ u(t) = 35 - 27e^{t\ln(20/27)/3} }\) and \(\displaystyle{ u(5) = 35 - 27e^{5\ln(20/27)/3} }\)

Final Answer

$\displaystyle{ u(t) = 35 - 27e^{t\ln(20/27)/3} }$ and $\displaystyle{ u(5) = 35 - 27e^{5\ln(20/27)/3} }$

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A 40^oF roast is placed in a 350^oF oven. After 2 hours, the temperature of the roast is 125^oF. Assuming the temperature of the roast follows Newton's Law of Cooling(Warming), find a formula for the temperature of the roast T as a function of its time in the oven, t, in hours. The roast is done when the internal temperature reaches 165^oF. When will the roast be done?

Problem Statement

Final Answer

$ u(t) = 350 - 310e^{kt} $ ^oF where $\displaystyle{ k = (1/2)\ln\left[ \frac{45}{62} \right] }$ and
$\displaystyle{ t = \frac{2\ln(37/62)}{\ln(45/62)} }$ hours

Problem Statement

Solution

Even though the temperature is increasing, we can still use the same equation. So we have $u(t) = T + (u_o - T) e^{kt}$ where the initial temperature $u_o = 40$. We are given the surrounding temperature as $T=350$, so we use the 2 hour point of $u(2)=125$ to determine $k$.

$ u(2) = 125 = 350 + (40-350)e^{2k} $

$ -225 = -310e^{2k} $

$\displaystyle{ \frac{45}{62} = e^{2k} }$

$\displaystyle{ \ln\left[ \frac{45}{62} \right] = 2k }$

$\displaystyle{ (1/2)\ln\left[ \frac{45}{62} \right] = k }$

So the equation of the temperature as a function of time is $ u(t) = 350 - 310e^{kt} $ where $\displaystyle{ k = (1/2)\ln\left[ \frac{45}{62} \right] }$.

Now to find the time when the roast will be done, we just let $u(t) = 165$ in this last equation and solve for $t$.

$ 165 = 350 - 310e^{kt} $

$ -185 = -310e^{kt} $

$ 37/62 = e^{kt} $

$ \ln(37/62) = kt $

$ (1/k)\ln(37/62) = t $

$\displaystyle{ \frac{\ln(37/62)}{(1/2)\ln(45/62)} = t }$

The exact answers are
$ u(t) = 350 - 310e^{kt} $ ^oF where $\displaystyle{ k = (1/2)\ln\left[ \frac{45}{62} \right] }$ and
$\displaystyle{ t = \frac{2\ln(37/62)}{\ln(45/62)} }$ hours
If your instructor wants you to give an approximate answer, we plug these numbers into a calculator to get $ u(t) \approx 350 - 310e^{-0.1602t} $^oF
$ t \approx 3.22 $ hours

Final Answer
\( u(t) = 350 - 310e^{kt} \) ^oF where \(\displaystyle{ k = (1/2)\ln\left[ \frac{45}{62} \right] }\) and \(\displaystyle{ t = \frac{2\ln(37/62)}{\ln(45/62)} }\) hours

Final Answer

$ u(t) = 350 - 310e^{kt} $ ^oF where $\displaystyle{ k = (1/2)\ln\left[ \frac{45}{62} \right] }$ and
$\displaystyle{ t = \frac{2\ln(37/62)}{\ln(45/62)} }$ hours

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Practice 2482
Solution

If you put a can of soda in the refrigerator with a temperature of 34 degrees (F). Suppose the initial temperature of the soda is 75 degrees and after 5 minutes, its temperature is 65 degrees. What will be the temperature of the soda in 30 minutes? When will the soda be 36 degrees?

Problem Statement

Solution

2482 video

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Practice 2483
Solution

A tofu turkey is taken from an oven when its temperature has reached 165F and is placed on a table in a room where the temperature is 78F. If the temperature of the turkey is 142F after half an hour, what is its temperature after 40 minutes? When will the turkey cool to 95F?

Problem Statement

Solution

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Practice 2484
Solution

The air temperature is 5 degrees C. Water cools from 100C to 75C in 5 minutes. How long will it take for the water to cool to 10C?

Problem Statement

The air temperature is 5 degrees C. Water cools from 100C to 75C in 5 minutes. How long will it take for the water to cool to 10C?

Solution

2484 video

video by AlRichards314

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A roast was taken out of a hardwood smoker when its' internal temperature had reached 180^oF and it was allowed to rest in a 75^oF house for 20 minutes after which its' internal temperature had dropped to 170^oF. Assume that the temperature of the roast follows Newton's Law of Cooling. a. Express the temperature (in ^oF) as a function of time t (in minutes). b. Find the time at which the temperature of the roast will dropped to 140^oF.

Problem Statement

Final Answer

a. $u(t) = 75+105e^{(t/20)\ln(19/21)}$^oF
b. $\displaystyle{ t = 20 \frac{\ln(13/21)}{\ln(19/21)} \approx 95.83 }$ min

Problem Statement

Solution

First, let's list what we are given in the problem statement. We know that the initial temperature is 180^oF. So $u(0) = u_o = 180$.
Next, we know that after 20 minutes, u(20) = 170 and the surrounding temperature is 75^oF, so $T = 75$.
Let's set up the equation and see what we have so far.
In general, $ u(t) = T + (u_o-T)e^{kt} $. Plugging in the given data, we have $ u(t) = 75 + (180-75)e^{kt} $.
a. We need to use the information $u(20)=170$ to determine k. Then we will have our answer to this part.

$ 170 = u(20) = 75+105e^{20k}$

$ 95 = 105e^{20k}$

$ 95/105 = e^{20k}$

$ 19/21 = e^{20k}$

$ \ln(19/21) = 20k $

$k = \ln(19/21)/20$

$u(t) = 75+105e^{(t/20)\ln(19/21)}$

b. Now we need to calculate the time at which the temperature drops to 140^oF. To do this we set the resulting equation from part a equal to 140 and solve for t.

$ 140 = 75+105e^{(t/20)\ln(19/21)} $

$ 65 = 105e^{(t/20)\ln(19/21)} $

$ 65/105 = e^{(t/20)\ln(19/21)} $

$ 13/21 = e^{(t/20)\ln(19/21)} $

$ \ln(13/21) = (t/20)\ln(19/21) $

$\displaystyle{ 20 \frac{\ln(13/21)}{\ln(19/21)} = t }$

To check that our answer is reasonable, we plug the last result into a calculator to get $t \approx 95.83 $ min, which seems reasonable.

Final Answer

a. $u(t) = 75+105e^{(t/20)\ln(19/21)}$^oF
b. $\displaystyle{ t = 20 \frac{\ln(13/21)}{\ln(19/21)} \approx 95.83 }$ min

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It takes 12 minutes for an object at 100C to cool to 80C in a room at 50C. How much longer will it take for it's temperature to decrease to 70C?

Problem Statement

It takes 12 minutes for an object at 100C to cool to 80C in a room at 50C. How much longer will it take for it's temperature to decrease to 70C?

Final Answer

$ t = 9.52 $ minutes

Problem Statement

It takes 12 minutes for an object at 100C to cool to 80C in a room at 50C. How much longer will it take for it's temperature to decrease to 70C?

Solution

2914 video

Final Answer

$ t = 9.52 $ minutes

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Find the time when an object reaching $80^o$F if it’s initial temperature is $150^o$F, the surrounding temperature is $70^o$F. We know that at $t=20$, it’s temperature $110^o$F.

Problem Statement

Find the time when an object reaching $80^o$F if it’s initial temperature is $150^o$F, the surrounding temperature is $70^o$F. We know that at $t=20$, it’s temperature $110^o$F.

Final Answer

$ t = 60 $ minutes

Problem Statement

Find the time when an object reaching $80^o$F if it’s initial temperature is $150^o$F, the surrounding temperature is $70^o$F. We know that at $t=20$, it’s temperature $110^o$F.

Solution

2915 video

video by Michel vanBiezen

Final Answer

$ t = 60 $ minutes

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$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$		$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$
$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$		$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$
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$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$		$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$
$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$		$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$
$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$		$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$

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$\int{\sin(x)~dx} = -\cos(x)+C$		$\int{\cos(x)~dx} = \sin(x)+C$
$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$		$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$
$\int{\sec(x)~dx} = $ $ \ln\abs{\sec(x)+\tan(x)}+C$		$\int{\csc(x)~dx} = $ $ -\ln\abs{\csc(x)+\cot(x)}+C$

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\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)		\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)		\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)		\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)		\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)		\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)		\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

\(\int{\sin(x)~dx} = -\cos(x)+C\)		\(\int{\cos(x)~dx} = \sin(x)+C\)
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)		\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)		\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

17Calculus Precalculus - Newton's Law of Cooling

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Michel vanBiezen - PreCalculus - Logarithmic & Exponential Functions (16 of 20) Newton's Law of Cooling [4min-55secs]

2482 video

2483 video

2484 video

2914 video

2915 video

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