Hi Aspirant!

It is given that the pan cools down from 94 degrees to 86 degree celcius in 2 minutes and the room temperature is 20 degree celcius.

So, T1 = 94 degrees

T2 = 86 degrees

T3 = 20 deg.

t' = 2 min (94 to 86 degrees)

t = ? (from 71 to 69 degrees)

So, as we know:

ln [ (T2-T3)/(T1-T3)] = - kt

=> ln [ ( 86 - 20) / ( 94-20) = -2k

after solving, we get k = - (1/2) ln ( 33/31)

Then , to find 't':

ln [(69-20)/ (71-20)] = -(1/2) ln ( 33/31) t

we get, t = [ln(49/51)] / [-(1/2)  ln ( 33/31)]

=> t = 0.69 minutes

=> t = 0.69 x 60 = 41.4 seconds

So, it will take 41.4 seconds to cool down the pan from 71 degrees to 69 degrees celcius.

Thankyou!