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Wize University Calculus 1 Textbook > Applications of Differentiation for Science

Newton’s Law of Cooling

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Newton’s Law of Cooling


The rate of cooling of an object, dTdt\displaystyle\frac{dT}{dt} ,is proportional to the temperature difference between the object and its surrounding (T−Ts)(T − T_s).

Newton’s Law of Cooling

dTdt=k(T−Ts)\boxed{\frac{dT}{dt}=k(T-T_s)}

where kk is the constant of proportionality,TT is the temperature of the object at time tt,and TsT_s is the temperature of the surroundings. The equation reduces to a growth/decay differential equation and the solution is
T(t)=(T(0)−Ts)e−kt+Ts\boxed{T(t)=(T(0)-T_s)e^{-kt}+T_s}

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Example: Newton’s law of cooling

A cupcake is removed from a −20∘C-20^{\circ}C freezer into a room at 20∘C20^{\circ}C. The unfortunate customer is distracted, letting the temperature of the delicious treat rise to a temperature of 0∘C0^{\circ}C in 30 minutes. If the optimal temperature of consumption is −10∘C-10^{\circ}C, and this system follows Newton’s law of cooling, when should he have consumed the treat?

T(0)=−20 T(0) = -20
Ts=20 T_s = 20
T(30)=0T(30)=0
T(?)=−10T(?)=-10

T(t)=(−20−20)e−kt+20=−40e−kt+20 T(t) = (-20-20)e^{-kt}+20 = -40e^{-kt}+20

T(30)=0=−40e−30k+20T(30) = 0 = -40e^{-30k}+20

k=−130ln⁡(1/2) k = -\dfrac{1}{30}\ln(1/2)

then

T(t)=−40e(1/30)ln⁡(1/2)t+20\displaystyle T(t)=-40e^{(1/30)\ln(1/2)t}+20

⇒−10=−40e(1/30)ln⁡(1/2)t+20\displaystyle \Rightarrow -10 = -40e^{(1/30)\ln(1/2)t}+20

⇒ln⁡(34)=130ln⁡(12)t\Rightarrow \ln\left(\dfrac{3}{4}\right) = \dfrac{1}{30}\ln\left(\dfrac{1}{2}\right)t

⇒t=30ln⁡(3/4)ln⁡(1/2)≈12.5 mins\Rightarrow t = 30\dfrac{\ln(3/4)}{\ln(1/2)}\approx \boxed{12.5 \text{ mins}}

Consider Newton's law of cooling
dTdt=−2+T10.\frac{\text{d}T}{\text{d}t}=-2+\frac{T}{10}.
Determine T(t)    if  T(0)=100T(t)\,\,\,\, \text{if } \, T(0)=100.



Extra Practice
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's Law of Cooling
A customer is served a cup of tea with temperature 60o C at a coffee shop. The shop has
room temperature 24o C. After 10 minutes, the temperature of the tea decreases to 54oC.
Using the (Newton's) law of cooling, determine the temperature after 20 minutes.
Newton's Law of Cooling, Rate Given
A liquid has temperature 100∘C100^{\circ}C in a room with temperature 20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of 1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Suppose a boiling cup of coffee at 100oC100^oC is allowed to sit in a room of temperature 25oC25^oC. You notice that a minute later, the coffee is 70oC 70^oC. After letting it sit out for 5 minutes from when it was boiling hot, you decide to put in cream, and the temperature decreases by a half. What is the temperature of the coffee now?
An object is in an environment of 30oC30 ^oC. It cools down from 100oC100^oC to 80oC80^oC in 5 minutes. What is the temperature after 8 minutes?
Newton's Law of Cooling
The body temperature T(t)T(t) of a butcher is usually 37.5 degrees Celsius. One day however, he locks himself in the meat freezer until he manages to break out twenty minutes later. The temperature of the freezer is normally -10 degrees Celsius. If body temperature follows Newton's Law of Cooling and he breaks out with body temperature 36 degrees Celsius, how long would he have to be in the freezer for his body temperature to reach 35 degrees Celsius?
Newton's law of cooling
Consider Newton's law of cooling
dTdt=2−T10. \frac{\text{d}T}{\text{d}t}=2-\frac{T}{10}.
Determine T(t)    if T(0)=100T(t)\,\,\,\, if\, T(0)=100.
Newton's Law of Cooling
A liquid has temperature 100∘C100^{\circ}C in a room with temperature
20∘C20^{\circ}C. At temperature 80∘C80^{\circ}C, the liquid is cooling at a rate of
1∘C1^\circ C per minute. When does the liquid reach 80∘C80^{\circ} C?
Newton's law of cooling
A beverage is put into an environment with a temperature of 200∘C.200^{\circ}C.The temperature of the liquid reads 20∘20^{\circ} at the beginning, and 30∘C30^{\circ}C after 20 minutes. How many minutes from the beginning does it take to reach 80∘C80^{\circ}C?
Newton's law of heating and cooling
Let’s learn how to cook pulled pork. A pork shoulder is removed from a refrigerator at 0∘C0^{\circ}C and placed into an already-hot slow cooker. The cooker initially cooks the pork for 2 hours on HIGH (150∘C)(150^{\circ}C), after which it is switched to LOW (100∘C)(100^{\circ}C) for another 8 hours. Assume the temperature of the cooker switches instantaneously and the pork follows Newton’s law of heating.
If the pork’s internal temperature is measured to be 70∘C70^{\circ}C at the end of the 8 hours, what is the proportionality constant for LOW cooking?
Newton's Law of heating and cooling
Let’s learn how to cook pulled pork. A pork shoulder is removed from a refrigerator at 0∘C0^{\circ}C and placed into an already-hot slow cooker. The cooker initially cooks the pork for 2 hours on HIGH (150∘C)(150^{\circ}C), after which it is switched to LOW (100∘C)(100^{\circ}C) for another 8 hours. Assume the temperature of the cooker switches instantaneously and the pork follows Newton’s law of heating.
If the pork is at 30∘C30^{\circ}C after 1 hour, what is its internal temperature when the cooker is switched to LOW?