A bowl filled with very hot soup cools from 98 ° C to 86^ C in 2...

From Newton's law of cooling, we have

$\dfrac{\DeltaQ}{\Deltat} = - K\left({T}_{avg} - {T}_{0}\right)$

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

Case $1:$

$∆ t = 2minutes, {T}_{0} = 22 ° C$

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

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

Case $2:$

$\dfrac{ms \times 6}{\Deltat} = - K\left(\dfrac{75 + 69}{2} - 22\right)$

$\dfrac{6}{\Deltat} = - \dfrac{K}{ms} ( 50 ) . . . \left(2\right)$

Dividing (2) by (1), we get

$⇒ \dfrac{6}{\Deltat ( 6 )} = \dfrac{50}{70}$

$⇒ \Deltat = \dfrac{7}{5} = 1 . 4min$