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Question: The temperature of a body in a room is \(100{}^\circ F\) . After five minutes, the temperature of th...

The temperature of a body in a room is 100∘F100{}^\circ F . After five minutes, the temperature of the body becomes 50∘F50{}^\circ F . After another 5 minutes, the temperature becomes 40∘F40{}^\circ F . What is the temperature of surroundings?

Explanation

Solution

Hint: We should know Newton's Cooling Law which is given as dtdT∝(T−S)\dfrac{dt}{dT}\propto \left( T-S \right) . After this, we have to do integration on both sides and the equation will be getting as log⁡(T−S)=−kt+c\log \left( T-S \right)=-kt+c . Then we have to find the constant term c by putting T as 100∘F100{}^\circ F at t=1t=1 . Then on getting the final equation we will put values given in question and then will make k as subject variable and equate both the equation. From this, we will get the surrounding temperature.

Complete step-by-step answer:
Here, we will use the formula according to Newton's Cooling Law given as dtdT∝(T−S)\dfrac{dt}{dT}\propto \left( T-S \right) assuming at time t and temperature of a body as T.
As the temperature of body is decreasing continuously, we will assume proportionality constant as negative i.e. −k-k . So, the equation will be written as dtdT=−k(T−S)\dfrac{dt}{dT}=-k\left( T-S \right) .
On rearranging the terms, we get dt(T−S)=−kdt\dfrac{dt}{\left( T-S \right)}=-kdt
Now, we will integrate on both sides of equation we will get
∫dt(T−S)=∫−kdt\int{\dfrac{dt}{\left( T-S \right)}}=\int{-kdt}
We know that ∫1a=log⁡a\int{\dfrac{1}{a}=\log a} so using, this we will get
log⁡(T−S)=−kt+c\log \left( T-S \right)=-kt+c ………………………(1)
Where c is a constant term.
So, here taking T as 100∘F100{}^\circ F at t=1t=1 we get
log⁡(100−S)=−k+c=c\log \left( 100-S \right)=-k+c=c ………………..(2) (considering k as constant so, writing it as one term c)
So, putting equation (2) in equation (1), we get
log⁡(T−S)=−kt+log⁡(100−S)\log \left( T-S \right)=-kt+\log \left( 100-S \right) ……………………….(3)
Now taking t=5t=5 and T as 50∘F50{}^\circ F and putting in equation (3), we get
log⁡(50−S)=−5k+log⁡(100−S)\log \left( 50-S \right)=-5k+\log \left( 100-S \right) ………………………(4)
Similarly, taking t=10t=10 and T as 40∘F40{}^\circ F and putting in equation (3), we get
log⁡(40−S)=−10k+log⁡(100−S)\log \left( 40-S \right)=-10k+\log \left( 100-S \right) …………………(5)
From equation (4), we will make constant term k as subject and on rearranging the terms, we get
⇒15log⁡(50−S)(100−S)=−k\Rightarrow \dfrac{1}{5}\log \dfrac{\left( 50-S \right)}{\left( 100-S \right)}=-k ……………………….(6)
Similarly, we will do in equation (5) making k as subject we get as
⇒110log⁡(40−S)(100−S)=−k\Rightarrow \dfrac{1}{10}\log \dfrac{\left( 40-S \right)}{\left( 100-S \right)}=-k …………………..(7)
Now, equating equation (6) and (7), we get
⇒110log⁡(40−S)(100−S)=15log⁡(50−S)(100−S)\Rightarrow \dfrac{1}{10}\log \dfrac{\left( 40-S \right)}{\left( 100-S \right)}=\dfrac{1}{5}\log \dfrac{\left( 50-S \right)}{\left( 100-S \right)}
On solving, we get
⇒log⁡(40−S)(100−S)=2log⁡(50−S)(100−S)\Rightarrow \log \dfrac{\left( 40-S \right)}{\left( 100-S \right)}=2\log \dfrac{\left( 50-S \right)}{\left( 100-S \right)}
So, removing the log from both the sides we get
⇒(40−S)(100−S)=((50−S)(100−S))2\Rightarrow \dfrac{\left( 40-S \right)}{\left( 100-S \right)}={{\left( \dfrac{\left( 50-S \right)}{\left( 100-S \right)} \right)}^{2}} (using the property nlog⁡m=mnn\log m={{m}^{n}} )
On taking LCM on LHS side and simplifying the terms we get
⇒(40−S)(100−S)=(50−S)2\Rightarrow \left( 40-S \right)\left( 100-S \right)={{\left( 50-S \right)}^{2}}
⇒4000−140S+S2=2500−100S+S2\Rightarrow 4000-140S+{{S}^{2}}=2500-100S+{{S}^{2}}
⇒4000−2500=140S−100S\Rightarrow 4000-2500=140S-100S
⇒1500=40S\Rightarrow 1500=40S
⇒150040=S=37.5∘\Rightarrow \dfrac{1500}{40}=S=37.5{}^\circ
Thus, the temperature of the surroundings is 37.5∘F37.5{}^\circ F .

Note: Remember that here temperature is decreasing at an interval of 5 minutes. So, negative sign will come in proportionality constant in case if plus sign will be there no change in answer. Also, in the line given as “After another 5 minutes, the temperature becomes 40∘F40{}^\circ F” here do not take it as 5. Here is a consecutive decrease of 5minutes when temperature was 50degree. So, time will be t=5+5=10t=5+5=10 . Instead of this if t equals 5 is taken then there will be change in the answer. So, don’t make this mistake.