An object is cooled from `75^(@)C` to `65^(@)C` in 2 min in a room at `30^(@)C`. The time taken to cool the same object from `55^(@)C` to `45^(@)C` in the same room is

Options

A. 5 minutes
B. 3 minutes
C. 4 minutes
D. 2 minutes

Answer

C

Explanation

From Newton's law of cooling,
`(dT)/(dt)=-k(T-T_(0))`
`implies int_(75)^(65)(dT)/(T-T_(0))= -kint_(0)^(120)dt`
Here, `T_(0)=30^(@)C`
`ln((65-30)/(75-30))= - k(120-0)`
`k = 2.09xx10^(-3)`
To cool from `55^(@)C` to `45^(@)C`
`int_(55)^(45)(dT)/(T-30)=-(2.09xx10^(-3))int_(t1)^(t2)dt`
`implies In((45-30)/(55-30))=-(2.09xx10^(-3))(t_(2)-t_(1))`
`implies t_(2)-t_(1)=244`
`approx4 min`