MHB Sammy's question at Yahoo Answers regarding Newton's Law of Cooling

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Newton's Law of Cooling describes how the temperature of an object changes in relation to the surrounding medium, stating that the rate of temperature change is proportional to the temperature difference. In the scenario presented, a hot coal at 160°C is placed in water at 5°C, and after 40 seconds, its temperature drops to 70°C. Using the provided data, the temperature of the coal after 1 minute is calculated to be approximately 47.09°C. Additionally, the time it takes for the coal to reach 20°C is determined to be about 107.49 seconds. This analysis effectively applies the principles of differential equations to solve the temperature change over time.
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Here is the question:

First order differential equation problem...?

rate of change of temperature between an object and its surrounding medium is proportional to the temperature difference. A hot coal of 160 degree celsius is immersed in water of 5 degree celsius. After 40 seconds, the temperature of coal is 70 degree celsius. Assume water is kept at constant temperature. Find
(a) the temperature of coal after 1 minute
(b) When does the temperture of coal reach 20 degree celsius

I have posted a link there to this topic so the OP can find my work.
 
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Re: sammy's question at Yahoo Answers regarding Newton's Law of Cooling

Hello sammy,

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:

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

The ODE is separable and may be written:

$$\frac{1}{T-M}\,dT=-k\,dt$$

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

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

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

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

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

Hence, we find:

(2) $$T(t)=\left(T_0-M \right)\left(\frac{T_1-M}{T_0-M} \right)^{\frac{t}{t_1}}+M$$

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

We may now use the given data to answer the questions:

$$T_0=160,M=5,t_1=40,T_1=70$$

(a) Using (2) and $t=60$, we find:

$$T(60)=\left(160-5 \right)\left(\frac{70-5}{160-5} \right)^{\frac{60}{40}}+5=\left(155 \right)\left(\frac{13}{31} \right)^{\frac{3}{2}}+5=5\left(1+13\sqrt{\frac{13}{31}} \right)\approx47.09244817717765$$

(b) Using (3) and $T(t)=20$, we find:

$$t=\frac{40\ln\left(\frac{20-5}{160-5} \right)}{\ln\left(\frac{70-5}{160-5} \right)}=\frac{40\ln\left(\frac{3}{31} \right)}{\ln\left(\frac{13}{31} \right)}\approx107.49243770293884$$
 
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