QJEEM25Ph52a7: JEE Main 2025 Physics - Class 11 JEEM25Ph52a7 Solution: A cup of coffee cools from 90°C to 80°C in t minutes when the…

Here, we will use the Newton's law of cooling.

We know, ${{{T_1} - {T_2}} \over {\Delta t}} = k\left( {{{{T_1} + {T_2}} \over 2} - {T_s}} \right)$ .... (1)

where, T $_1$ and T $_2$ are initial and final temperatures respectively.

$\mathrm{T_s}$ = surrounding temperature

$\Delta t$ = time taken

$k$ = positive constant

Given,

Ist case :

${T_1} = 90^\circ C$

${T_2} = 80^\circ C$

$\Delta t = t$ , ${T_s} = 20^\circ C$

IInd case :

${T_1} = 80^\circ C$

${T_2} = 60^\circ C$

${T_s} = 20^\circ C$ , $\Delta t = ?$

Now, for Ist case,

${{90 - 80} \over t} = k\left( {{{90 + 80} \over 2} - 20} \right)$ ...... (from (1))

$\Rightarrow {{10} \over t} = k\left( {85 - 20} \right)$

$\Rightarrow {{10} \over t} = 65k$ .... (2)

For IInd case,

${{80 - 60} \over {\Delta t}} = k\left( {{{80 + 60} \over 2} - 20} \right)$ .... (From (1))

$\Rightarrow {{20} \over {\Delta t}} = k\left( {70 - 20} \right)$

$\Rightarrow {{20} \over {\Delta t}} = 50k$ .... (3)

Now, by (2) $\div$ (3),

${{{{10} \over t}} \over {{{20} \over {\Delta t}}}} = {{13} \over {10}}$

$\Rightarrow {{\Delta t} \over t}\left( {{1 \over 2}} \right) = {{13} \over 5}$

$\Rightarrow \Delta t = {{13} \over 5}t$