A body cools from `50^@C` to `49^@C` in 5 s. How long will it take to cool from `40^@C` to `39.5^@C`? Assume the temperature of surroundings to be `30^@C` and Newton's law of cooling to be valid:

Options

A. `2.5s`
B. `10s`
C. `20s`
D. `5s`

Answer

B

Explanation

From Newton's law of cooling,
`(theta_(1)-theta_(2))/t prop((theta_(1)+theta-(2))/2-theta_(0))`
where `theta_(0)=` temperature of surroundings.
`:.(50-49)/(t_(1))prop ((50-49)/2-30)`………….i
and `(40-39)/(t_(2))prop((40+39)/2-30)`………ii
On dividing Eq. (ii) by Eq. (i) we get
`(t_(2))/(t_(1))=39/19`

`impliest_(2)=39*19xx5`
`=10s`