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Question: The temperature \(T\) of a cooling object drops at a rate proportional to the difference \((T-S)\) w...

The temperature TT of a cooling object drops at a rate proportional to the difference (T−S)(T-S) where SS is the constant temperature of the surrounding medium. If initially T=150∘CT=150{}^\circ C, find the temperature of the cooling object at any time tt.

Explanation

Solution

We are given the rate of change of temperature of a body. Using that we can form a differential equation by adding a positive proportionality constant. We can integrate the differential equation to get the required equation of temperature at any time. We can use the given initial condition to eliminate the constant of integration to get the required answer. Try it, you will definitely get the answer.

Complete step by step solution:
We are given that the rate of cooling of a body with temperature TT is proportional to (T−S)(T-S). We can write this in differential form.
dTdt∝−(T−S)\dfrac{dT}{dt}\propto -(T-S)
We put the negative sign here because the temperature is decreasing.
Now, let k>0k>0 be the proportionality constant. Then we get the differential equation as,
dTdt=−k(T−S)\dfrac{dT}{dt}=-k(T-S)
So, let us solve the equation using the method of separation of the variable.
dT(T−S)=−kdt\dfrac{dT}{(T-S)}=-kdt
Now as we have same variable on same side of equation, so now integrating both sides we get,
∫dT(T−S)=−∫kdt\int{\dfrac{dT}{(T-S)}}=-\int{kdt}
We know that, ∫dxx−a=log⁡(x−a)\int{\dfrac{dx}{x-a}=\log (x-a)} and ∫dx=(x)\int{dx=(x)}.
So, applying above to the equation we get,
log⁡(T−S)=−kt+c\log (T-S)=-kt+c ………….. (1)
On applying initial condition that, T=150∘CT=150{}^\circ C at t=0t=0, we get,
log⁡(150−S)=−k×0+c\log (150-S)=-k\times 0+c
c=log⁡(150−S)c=\log (150-S) ……… (2)
Now substituting the value of cc in (1) we get,
log⁡(T−S)=−kt+log⁡(150−S)\log (T-S)=-kt+\log (150-S)
Now rearranging the equation we get,
log⁡(T−S)−log⁡(150−S)=−kt\log (T-S)-\log (150-S)=-kt
Now we know the property of log that, log⁡a−log⁡b=log⁡(ab)\log a-\log b=\log \left( \dfrac{a}{b} \right).
Applying above property we get,
log⁡(T−S150−S)=−kt\log \left( \dfrac{T-S}{150-S} \right)=-kt
Now taking exponents on both sides we get,
elog⁡(T−S150−S)=e−kt{{e}^{\log \left( \dfrac{T-S}{150-S} \right)}}={{e}^{-kt}}
We also know that, elog⁡a=a{{e}^{\log a}}=a.
Applying above property we get,
(T−S150−S)=e−kt\left( \dfrac{T-S}{150-S} \right)={{e}^{-kt}}
Now cross multiplying we get,
T−S=(150−S)e−ktT-S=(150-S){{e}^{-kt}}
Now simplifying in simple manner we get,
T=S+(150−S)e−ktT=S+(150-S){{e}^{-kt}}

Therefore, the temperature of cooling body at time tt is given by T=S+(150−S)e−ktT=S+(150-S){{e}^{-kt}} where SS is temperature of surrounding medium and k>0k>0 is the constant of proportionality.

Note:
We have used the concept of differential equations to represent the rate of cooling. We are given the rate of proportionality where kk is proportionality constant. We have taken k>0k>0 if we took k=0k=0 then the rate of change will be zero and if k<0k<0, the rate of change will reverse.