A liquid takes `10` minutes to cool from `80^(@)C` to `50^(@)C`. The temperature of the surroudings is `20^(@)C`. Assuming that the Newton's law of cooling is obeyed, the cooling constant will be -

Options

A. `0.056//mt`
B. `0.042//mt`
C. `0.081//mt`
D. `0.069//mt`

Explanation

`(DeltaT)/(Deltat)=k(theta_(w)-theta_(s))`
`(80-50)/(10 min)=k((80+50)/(2)-20)`
`3=k(65-20)`
`k=(1)/(15)=.063`