A cup of coffee cools from 90 ° C to 80 ° C in tmin , when the room...

From Newton's Law of cooling,

$\dfrac{\DeltaT}{\Deltat} = k\left(T - {T}_{s}\right)$

where,

$⇒ \dfrac{{T}_{2} - {T}_{1}}{t} - k\left(\dfrac{{T}_{1} + {T}_{2}}{2} - {T}_{s}\right){T}_{s}$

Where, ${T}_{1} =$ Initial temperature of body

${T}_{2} =$ Final temperature of body

${T}_{s} =$ surrounding temperature

$t =$ time taken

For case$1,$

$\dfrac{90 - 80}{t} = k\left(\dfrac{90 + 80}{2} - 20\right)$

$⇒ \dfrac{10}{t} = 65k$

$⇒ k = \dfrac{10}{65t} \ldots \left(i\right)$

For case $2,$

$\dfrac{80 - 60}{{t}^{'}} = k\left(\dfrac{80 + 60}{2} - 20\right)$

$⇒ \dfrac{20}{{t}^{'}} = 50k$

Put the value $k$ from eq.\left(\text{i}\right)$,</p><p>$⇒ \dfrac{20}{{t}^{'}} = 50 \times \dfrac{10}{65t}$</p><p>$⇒ {t}^{'} = \dfrac{20 \times 65}{500}t$</p><p>$⇒ {t}^{'} = \dfrac{13}{5}t