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A liquid in a beaker has temperature phi...

A liquid in a beaker has temperature `phi(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

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
B
`(d theta)/(dt)=-K(theta-theta_(0))`
`int_(theta_(0))^(theta)(d theta)/(theta-theta_(0))=-Kint_(0)^(t)dt`
`lntheta_(0)-theta_(0)=-Kt+C`
The above equation shows that graph is a straight line Correct choice is (b).
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