# A body cools down, from to 50c, in 5min, and to 40c, in another 8min. Find the temperature of the...

## Question:

A body cools down, from to 50c, in 5min, and to 40c, in another 8min. Find the temperature of the surroundings.

## The Law of Cooling

The law of cooling is a way to calculate the temperature of an object as heat is transferred from the body into the surroundings. This law states that the rate of change in temperature is proportional to the difference between the object temperature and the ambient temperature ( from the surroundings). The formula is then given as,

$$\frac{dT}{dt}=-k(T_{ave}-T_s)$$

where

{eq}t {/eq} is the time

{eq}T {/eq} is the temperature of the body at any time t

{eq}T_s {/eq} is the temperature of the surroundings

{eq}T_i {/eq} is the initial temperature of the body, and

{eq}k {/eq} is the constant of proportionality

For the sake of convenience, let us separate the two instances where the temperature drops.

Part 1. For the first part, we set the following to be:

• Surrounding temperature: {eq}\begin{align*} T_s \end{align*} {/eq}
• The initial temperature: {eq}\begin{align*} T_i \end{align*} {/eq}
• The final temperature: {eq}\begin{align*} T_f = 50^o C \\ \end{align*} {/eq}
• Time of cooling down: {eq}\begin{align*} t=5~mins \end{align*} {/eq}

The average temperature is then calculated to be,

\begin{align*} T_{ave}&=\frac{T_f+T_i}{2}=\left( \frac{T_i+50}{2}\right)~^o C&\text{}\\\\ \end{align*}

The difference between the body temperature and the surrounding is then equal to

\begin{align*} T_{diff}&=T_{ave}-T_s=\left( \frac{T_i+50}{2}-T_s\right)~^o C&\text{}\\\\ \end{align*}

On the other hand, calculating the rate of temperature decrease, we have

\begin{align*} \frac{\Delta T}{t}&=\left( \frac{T_i-50}{5}\right)~\frac{^o C}{min} &\text{}\\\\ \end{align*}

Using Newton's law of cooling, we obtain the expression

\begin{align*} \frac{dT}{dt}&=-k\left( \frac{T_i+50}{2}-T_s\right)&\text{[Substitute the rate given in equation 3]}\\\\ &\boxed{\left( \frac{T_i-50}{5}\right)=-k\left( \frac{T_i+50}{2}-T_s\right)}&\text{}\\\\ \end{align*}

Part 2. Doing the same procedure in part 1 but with the following given:

• Surrounding temperature: {eq}\begin{align*} T_s \end{align*} {/eq}
• The initial temperature: {eq}\begin{align*} T_i=50 \end{align*} {/eq}
• The final temperature: {eq}\begin{align*} T_f = 40^o C \\ \end{align*} {/eq}
• Time of cooling down: {eq}\begin{align*} t=8~mins \end{align*} {/eq}

The average temperature is then calculated to be,

\begin{align*} T_{ave}&=\frac{T_f+T_i}{2}=\left( \frac{50+40}{2}\right)=45~^o C&\text{}\\\\ \end{align*}

The difference between the body temperature and the surrounding is then equal to

\begin{align*} T_{diff}&=T_{ave}-T_s=\left( 45-T_s\right)~^o C&\text{}\\\\ \end{align*}

On the other hand, calculating the rate of temperature decrease, we have

\begin{align*} \frac{\Delta T}{t}&=\left( \frac{50-40}{8}\right)=1.25~\frac{^o C}{min} &\text{}\\\\ \end{align*}

Using Newton's law of cooling, we obtain the expression

\begin{align*} \frac{dT}{dt}&=-k\left(45-T_s\right)&\text{[Subsitute the rate given in equation 7]}\\\\ &\boxed{1.25=-k\left(45-T_s\right)}&\text{}\\\\ \end{align*}

Part 3. To obtain the temperature of the surroundings, we divide equation 4 by equation 8. And so we have

\begin{align*} \frac{\left( \frac{T_i-50}{5}\right)}{1.25}&=\frac{-k\left( \frac{T_i+50}{2}-T_s\right)}{-k\left(45-T_s\right)}&\text{[Rearrange]}\\\\ -k\left(45-T_s\right)\left( \frac{T_i-50}{5}\right)&=-k\left( \frac{T_i+50}{2}-T_s\right)\left( 1.25\right)&\text{[Cancel out -k]}\\\\ \left(45-T_s\right)\left( \frac{T_i-50}{5}\right)&=\left( \frac{T_i+50}{2}-T_s\right)\left( 1.25\right)&\text{[Distribute the terms]}\\\\ 45T_i-2250-T_sT_i+50T_s&=5\left( 0.625T_i+31.25-1.25T_s\right)&\text{[Rearrange and combine like terms]}\\\\ 56.25T_s-T_sT_i&=-41.875T_i+2406.25&\text{[Factor out } T_s \text{]}\\\\ \left( 56.25-T_i\right)T_s&=2406.25-41.875T_i&\text{[Simplify]}\\\\ \therefore~ &\boxed{T_s=\frac{(2406.25-41.875T_i)}{\left( 56.25-T_i\right)}}&\text{[General formula with initial temperature unknown]}\\\\ \end{align*} 