A liquid in a beaker has temperature `theta(t)` at time t and `theta_0` is temperature of surroundings, then according to Newton's law of cooling the correct graph between `log_e( theta-theta_0)` and t is :

Options

A.
B.
C.
D.

Answer

A

Explanation

(a) Newton's law of cooling
`(d theta)/(dt)=-K( theta- theta_0)implies(d theta)/( theta- theta_0)-kdt`
Integrating
`implies log( theta- theta_0)=-kt+c`
Which represents an equation of straight line.
Thus the option (a) is correct.