Question

A body cools according to Newton’s law from ${{100}^{\circ }}C$ to ${{60}^{\circ }}C$ in 20 minutes. The temperature of the surrounding being ${{20}^{\circ }}C$. How long will it take to cool down to ${{30}^{\circ }}C$.

Answer 1

In this problem, we are given a body cools according to Newton’s law from ${{100}^{\circ }}C$ to ${{60}^{\circ }}C$ in 20 minutes. The temperature of the surrounding being ${{20}^{\circ }}C$. We should find the time taken by the body to cool down to ${{30}^{\circ }}C$. We can first write the equation of Newton’s law of cooling. We can then integrate them and substitute the required data to find the time.

Complete step by step answer:
Here we are given a body that cools according to Newton’s law from ${{100}^{\circ }}C$ to ${{60}^{\circ }}C$ in 20 minutes. The temperature of the surrounding being ${{20}^{\circ }}C$. We should find the time taken by the body to cool down to ${{30}^{\circ }}C$.
Let ${{\theta }^{\circ }}C$ be the temperature of the body at time t.
We know that according to Newton’s law of cooling, we can write the equation as,

& \Rightarrow \dfrac{d\theta }{dt}\propto \left( \theta -20 \right) \\\ & \Rightarrow \dfrac{d\theta }{dt}=-k\left( \theta -20 \right),k>0 \\\ \end{aligned}$$ We can now write the above step as, $$\Rightarrow \dfrac{d\theta }{\left( \theta -20 \right)}=-kdt$$ We can now integrate on both sides, we get $$\begin{aligned} & \Rightarrow \int{\dfrac{d\theta }{\left( \theta -20 \right)}}=-k\int{dt} \\\ & \Rightarrow \log \left( \theta -20 \right)=-kt+C.......(1) \\\ \end{aligned}$$ We know that initially we have t = 0, $$\theta ={{100}^{\circ }}C$$, substituting in (1), we get $$\begin{aligned} & \Rightarrow \log \left( 100-20 \right)=-k\left( 0 \right)+C \\\ & \Rightarrow C=\log 80......(2) \\\ \end{aligned}$$ We can substitute (2) in (1), we get $$\begin{aligned} & \Rightarrow \log \left( \theta -20 \right)=-kt+\log 80 \\\ & \Rightarrow \log \left( \dfrac{\theta -20}{80} \right)=-kt \\\ \end{aligned}$$ Now, when t = 20 and $$\theta ={{60}^{\circ }}$$, substituting in the above step, we get $$\begin{aligned} & \Rightarrow \log \left( \dfrac{60-20}{80} \right)=-k\times 20 \\\ & \Rightarrow k=-\dfrac{1}{20}\log \left( \dfrac{1}{2} \right).......(3) \\\ \end{aligned}$$ Now (1) becomes, $$\Rightarrow \log \left( \dfrac{\theta -20}{80} \right)=\dfrac{t}{20}\log \left( \dfrac{1}{2} \right)$$ We can now substitute $$\theta ={{30}^{\circ }}$$ in the above step, we get $$\begin{aligned} & \Rightarrow \log \left( \dfrac{30-20}{80} \right)=\dfrac{t}{20}\log \left( \dfrac{1}{2} \right) \\\ & \Rightarrow \log \left( \dfrac{1}{8} \right)=\log {{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{20}}} \\\ \end{aligned}$$ We can now write the above step as, $$\begin{aligned} & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{3}}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{20}}} \\\ & \Rightarrow 3=\dfrac{t}{20} \\\ & \Rightarrow t=60\min \\\ \end{aligned}$$ Therefore, the required time is 60 minutes. **Note:** We should always remember that the Newton’s law of cooling equation is $$\dfrac{d\theta }{dt}\propto \left( \theta -20 \right)$$, where we can take the proportion and add a constant equating the both sides. We should also know some of the integration formulas to solve these types of problems.