Nano's question at Yahoo Answers regarding Newton's Law of Cooling

In summary, the question asks how long Sandy must wait before opening her kiln after turning it off and the temperature gauge reads 700°F in 10 minutes. Using Newton's Law of Cooling, and the given data, we can solve for the time it takes for the kiln to cool to 155°F, which is approximately 146 minutes.
  • #1
MarkFL
Gold Member
MHB
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Here is the question:

I need help with this problem !?

Sandy manages a ceramics shop and uses a 800°F kiln to fire ceramic greenware. After turning off her kiln, she must wait until its temperature gauge reaches 155°F before opening it and removing the ceramic pieces. If room temperature is 70°F and the gauge reads 700°F in 10 minutes, how long must she wait before opening the kiln? Assume the kiln cools according to Newton's Law of Cooling.
(Round you anwer to the nearest whole minute)
a) 512 mins B)94 mins c)369 mins D)146 mins

I have posted a link there to this topic so the OP can see my work.
 
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  • #2
Hello Nano,

Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:

\(\displaystyle \frac{dT}{dt}=-k(T-M)\) where \(\displaystyle T(0)=T_0,\,0<k\in\mathbb{R}\) and \(\displaystyle T>M\).

The ODE is separable and may be written:

\(\displaystyle \frac{1}{T-M}\,dT=-k\,dt\)

Integrating, using the boundaries, and dummy variables of integration, we find:

\(\displaystyle \int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t v\,dv\)

\(\displaystyle \ln\left(\frac{T(t)-M}{T_0-M} \right)=-kt\)

If we know another point \(\displaystyle \left(t_1,T_1 \right)\) we may now find $k$:

\(\displaystyle -k=\frac{1}{t_1}\ln\left(\frac{T_1-M}{T_0-M} \right)\)

Hence, we find:

\(\displaystyle t=\frac{t_1\ln\left(\frac{T(t)-M}{T_0-M} \right)}{\ln\left(\frac{T_1-M}{T_0-M} \right)}\)

Using the following given data:

\(\displaystyle t_1=10,\,T(t)=155,\,M=70,\,T_0=800,\,T_1=700\)

we find:

\(\displaystyle t=\frac{10\ln\left(\frac{155-70}{800-70} \right)}{\ln\left(\frac{700-70}{800-70} \right)}=\frac{10\ln\left(\frac{17}{146} \right)}{\ln\left(\frac{63}{73} \right)}\approx145.9628332694268\)

Therefore d) is the correct answer.
 

FAQ: Nano's question at Yahoo Answers regarding Newton's Law of Cooling

What is Newton's Law of Cooling?

Newton's Law of Cooling is a physical law that describes the rate at which an object cools down or loses heat to its surroundings. It states that the rate of cooling of an object is proportional to the difference between the object's temperature and the temperature of its surroundings.

How is Newton's Law of Cooling applied in real life?

Newton's Law of Cooling is applied in various fields such as meteorology, engineering, and food preservation. It is used to predict the cooling rate of hot objects, design cooling systems, and determine the shelf life of perishable goods.

What is the formula for Newton's Law of Cooling?

The formula for Newton's Law of Cooling is dT/dt = -k(T - Ts), where dT/dt is the rate of change of temperature over time, k is the cooling constant, T is the temperature of the object, and Ts is the temperature of the surroundings.

How does the temperature difference between the object and its surroundings affect the cooling rate?

The larger the temperature difference between the object and its surroundings, the faster the cooling rate will be. This is because there is a greater temperature gradient, leading to a higher rate of heat transfer.

Can Newton's Law of Cooling be applied to objects that are heating up?

Yes, Newton's Law of Cooling can also be applied to objects that are heating up. In this case, the formula would be dT/dt = k(Ts - T), where Ts is the temperature of the surroundings and T is the temperature of the object. This is often used in thermodynamics to study heat transfer in various systems.

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