A body cools down from `60^@C` to `55^@C` in 30 s. Using newton's law of cooling calculate the time taken by same body to cool down from `55^@C` to `50^@C`. Assume that the temperature of surrounding is `45^@C`.

Options

A. 41.28 sS
B. 55.28 SS
C. 51.28 s
D. 60.28 S

Answer

C

Explanation

According to Newton's law of cooling,
`(T_(1)-T_(2))/t=K[(T_(1)+T_(2))/(2)-theta_(0)]`
Case I `(60-55)/(30)=K [(60+55)/(2)-45]` (i)1
Case II `(55-50)/(t)=K [(50+50)/(2)-45]` ….(ii)
Dividing Eq. (i)by Eq . (ii), we get,
`t=51.28 s`