From Newton's law of cooling, we have

$\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}=-K\left(T_{\mathrm{avg}}-T_{\mathit{0}}\right)$

Where, $Q=ms∆T$, $m$ is the mass, $s$ is the specific heat and $∆T$ is the temperature change.

Case $1:$

$∆\mathrm{t}=2 \mathrm{minutes}, {\mathrm{T}}_0=22°\mathrm{C}$

$\frac{ms\times 12}{2}=-K\left(\frac{98+86}{2}-22\right)$

$6=-\frac{K}{ms}\left(70\right) ...\left(1\right)$

Case $2:$

$\frac{ms\times 6}{\Delta t}=-K\left(\frac{75+69}{2}-22\right)$

$\frac{6}{\Delta t}=-\frac{K}{ms}\left(50\right) ...\left(2\right)$

Dividing (2) by (1), we get

$\Rightarrow \frac{6}{\Delta t\left(6\right)}=\frac{50}{70}$

$\Rightarrow \Delta t=\frac{7}{5}=1.4 \min$