A body is heated to \(110^{\circ} \mathrm{C}\) and placed in air at \(10^{\circ} \mathrm{C}\). After 1 hour its temperature is
\(60^{\circ} \mathrm{C}\). The additional time required for it to cool to \(30^{\circ} \mathrm{C}\) is

(D)
Let \(\theta\) be the temperature of body at time \(t\). Temperature of air is given to be \(10^{\circ} \mathrm{C}\). \(=\theta_{0}\) (say).
\(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto\left(\theta-\theta_{0}\right) \)
\( \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}\left(\theta-\theta_{0}\right), \mathrm{k}>0 \)
\( \therefore \int \frac{\mathrm{d} \theta}{\theta-\theta_{0}}=\int-\mathrm{kt} \quad \Rightarrow \log \left|\theta-\theta_{0}\right|=-\mathrm{kt}\) \(+\log \mathrm{c}...(1)\)
\(\therefore\left(\frac{\theta-\theta_{0}}{\mathrm{c}}\right)=\mathrm{e}^{-\mathrm{kt}}\) \(\therefore \theta=\theta_{0}+\mathrm{ce}^{-\mathrm{kt}}\) When \(\mathrm{t}=0, \theta=110\) and \(\theta_{0}=10, \quad \therefore \theta=10+100 \mathrm{e}^{-\mathrm{kt}}\) \(\quad 110=10+\mathrm{c} \Rightarrow \mathrm{c}=100 \quad \therefore \quad \mathrm{c}\) When \(\theta=60^{\circ} \mathrm{C}, \mathrm{t}=1\) Substituting value in equation (1), \(\therefore \quad \log \left(\frac{60-10}{100}\right)=-\mathrm{k} \times 1\)
\(\quad \mathrm{k}=-\log \left(\frac{1}{2}\right)\)
\(\therefore \quad \log \left(\frac{\theta-10}{100}\right)=\mathrm{t} \cdot \log \left(\frac{1}{2}\right)\)
When \(\theta=30^{\circ} \mathrm{C}\), then
\(\log \left(\frac{30-10}{100}\right)=\mathrm{t} \cdot \log \left(\frac{1}{2}\right)\)
\(\log \left(\frac{20}{100}\right)=t \cdot \log \frac{1}{2}\)
\(-\log 5=-t \log 2 \Rightarrow t=\frac{\log 5}{\log 2}\)
\(\therefore\) Additional time required is \(\frac{\log 5}{\log 2}-1\)