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a simple method for these types of problem is average temperature method ...
rate of cooling = dT/dt = k [(Ti+Tf)/2 - To]
To is temperature of surrounding ...
liquid takes 10minutes to cool from 70 to 50 so
rate = 70-50/10 = k [ (70+50)/2 -20 ]
2 = k[ 60-20]
k = 1/20 ...................1
let time taken by liquid to cool from 80 to 60 is t then
rate = dT/t = 20/t = k [ (80+60)/2 - 20]
20/t = 50k
t = 2/5k
k = 1/20 from eq 2 so
t = 8 mins
this is the required time , approve my ans if u like
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