Water at $100^{\circ} \mathrm{C}$ cools in 10 minutes to $80^{\circ} \mathrm{C}$ in a room temperature of…

Water at $100^{\circ} \mathrm{C}$ cools in 10 minutes to $80^{\circ} \mathrm{C}$ in a room temperature of $25^{\circ} \mathrm{C}$, then the temperature of water after 20 minutes will be
  1. $65.33^{\circ} \mathrm{C}$
  2. $69.33^{\circ} \mathrm{C}$
  3. $60.33^{\circ} \mathrm{C}$
  4. $63.33^{\circ} \mathrm{C}$

Solution

$\begin{aligned} & \frac{\mathrm{d} T}{\mathrm{~d} t}=-k(T-25) \\ & \Rightarrow T-25=e^{-k t+c} \\ & \Rightarrow T=25+e^c \cdot e^{-k t} \\ & \text { for } t=0, T=100 \\ & \Rightarrow e^c=75 \\ & \Rightarrow T=25+75 e^{-k t} \\ & \text { for } t=10, T=80 \\ & \Rightarrow 80=25+75 e^{-k \times 10} \\ & \Rightarrow-10 k=\log \left(\frac{11}{55}\right) \\ & \text { i.e., } T=25+75 e^{\left(\frac{1}{10} \log \frac{11}{15}\right) t} \\ & \text { Now for } t=20, T=25+75 e^{20 \times \frac{1}{10} \log \frac{11}{15}}=25+75 \times\left(\frac{11}{15}\right)^2 \\ & \Rightarrow T=25+40.33=65.33 \\ & \end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 1)

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