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

[MarkFL]

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.

 

[MarkFL]

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.