Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A liquid in a beaker has temperature at time t and <img alt="\Theta _{0}" src="https://learn.careers360.c

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}\left ( \Theta -\Theta _{0} \right ) and t is :

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

According to Newton’s law of cooling

\frac{d\Theta }{dt}=-k(\Theta -\Theta _{0})or\frac{d\Theta }{\Theta -\Theta _{0}}=-kdt

Integrating both sides, we get

\int \frac{d\Theta }{\Theta -\Theta _{0}}=\int -kdt

Log_{e}(\Theta -\Theta _{0})=-kt+C

where C is a constant of integration. So, the graph between Log_{e}(\Theta -\Theta _{0}) and t is a straight line with a negative slope. Option (a) represents the correct graph.

Posted by

Ritika Harsh

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2025 registration deadline extended to April 24; application fee, documents required

Latest Question