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. `55.26 s`
B. `51.28 s`
C. `81.28 s`
D. `85.26 s`

Answer

B

Explanation

Assume that a body cools down from temperature `theta_i` to `theta_f` in t seconds, and `theta_c` is the temperature of surrounding. Applying newton's law of cooling.
According to Newton's law of cooling
`(dtheta)/(dt)=-K(theta-theta_0)`,`theta_0=45^@C`,`int(d theta)/(theta-45)=-intdt` (K is constant)
From `t=0s` to `t=30s`, `theta` changes from `60^@C` to `55^@C`.
`int_(55)^(50)(d theta)/(theta-45)=-Kint_(30)^tdtimpliesIn((50-45)/(60-45))=-K(30-0)` ..(i)
`int_(55)^(50)(d theta)/(theta-45)=-Kint_(30)^tdtimpliesIn((50-45)/(60-45))=-K(t-30)` ..(ii)
Divide Eq. (ii) Eq. (i) set:
`(t-30)/(30)=(In(50-45)/(55-45))/(ln(55-45)/(60-45))implies(t-30)/(30)=(ln2)/(ln 3//2)`
`impliest=30(1+(ln2)/(ln3//2))=81.28s`
Time from `theta=55^@C` to `theta=50^@C`. is `(t-30)=(81.28-(30)=51.28s`.