A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

<p>Here, we will use the Newton's law of cooling.</p> <p>We know, $${{{T_1} - {T_2}} \over {\Delta t}} = k\left( {{{{T_1} + {T_2}} \over 2} - {T_s}} \right)$$ .... (1)</p> <p>where, T$_1$ and T$_2$ are initial and final temperatures respectively.</p> <p>$\mathrm{T_s}$ = surrounding temperature</p> <p>$\Delta t$ = time taken</p> <p>$k$ = positive constant</p> <p>Given,</p> <p>Ist case :</p> <p>${T_1} = 90^\circ C$</p> <p>${T_2} = 80^\circ C$</p> <p>$\Delta t = t$, ${T_s} = 20^\circ C$</p> <p>IInd case :</p> <p>${T_1} = 80^\circ C$</p> <p>${T_2} = 60^\circ C$</p> <p>${T_s} = 20^\circ C$, $\Delta t = ?$</p> <p>Now, for Ist case,</p> <p>${{90 - 80} \over t} = k\left( {{{90 + 80} \over 2} - 20} \right)$ ...... (from (1))</p> <p>$\Rightarrow {{10} \over t} = k\left( {85 - 20} \right)$</p> <p>$\Rightarrow {{10} \over t} = 65k$ .... (2)</p> <p>For IInd case,</p> <p>${{80 - 60} \over {\Delta t}} = k\left( {{{80 + 60} \over 2} - 20} \right)$ .... (From (1))</p> <p>$\Rightarrow {{20} \over {\Delta t}} = k\left( {70 - 20} \right)$</p> <p>$\Rightarrow {{20} \over {\Delta t}} = 50k$ .... (3)</p> <p>Now, by (2) $\div$ (3),</p> <p>${{{{10} \over t}} \over {{{20} \over {\Delta t}}}} = {{13} \over {10}}$</p> <p>$\Rightarrow {{\Delta t} \over t}\left( {{1 \over 2}} \right) = {{13} \over 5}$</p> <p>$\Rightarrow \Delta t = {{13} \over 5}t$</p>