If a body is heated to $110^{\circ} \mathrm{C}$ and placed in air at $10^{\circ} \mathrm{C}$ after 1 hour…

If a body is heated to $110^{\circ} \mathrm{C}$ and placed in air at $10^{\circ} \mathrm{C}$ after 1 hour its temperature is $60^{\circ} \mathrm{C}$, then the additional time required for it to cool to $30^{\circ} \mathrm{C}$ is
  1. $\frac{\log 5}{\log 2} \mathrm{hrs}$
  2. $\frac{\log 2}{\log 5} \mathrm{hrs}$
  3. $\left(\frac{\log 5}{\log 2}-1\right) \mathrm{hrs}$
  4. $\left(\frac{\log 5}{\log 2}+1\right) \mathrm{hrs}$

Solution

$\begin{aligned} & \frac{\mathrm{d} T}{\mathrm{~d} t}=-K(T-10) \\ & \Rightarrow T-10=e^{-k t+C} \\ & \Rightarrow T=10+e^C \cdot e^{-k t} \end{aligned}$ For $t=0$, $\begin{aligned} & T=110 \\ & \Rightarrow e^C=100 \text { i.e. } \\ & T=10+100 \cdot e^{-k t} \end{aligned}$ For $t=1$, $\begin{aligned} & T=60 \\ & \Rightarrow 60=10+100 \cdot e^{-k \times 1} \\ & \Rightarrow-k=\log \frac{1}{2} \\ & \Rightarrow k=\log 2 \end{aligned}$ $\Rightarrow T=10+100 \cdot e^{-\left(\log _2\right) t}$ Putting $T=30$ $\begin{aligned} & 30=10+100 \cdot e^{-\left(\log _2\right) t} \\ & \Rightarrow \log \left(\frac{1}{5}\right)=-(\log 2) t \\ & \Rightarrow t=\frac{\log 5}{\log 2} \end{aligned}$ Additional time $=t-1=\frac{\log 5}{log2}-1$

Asked in: MHT CET 2022 (11 Aug Shift 1)

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