Question

A body in air at 25℃ cools from 100℃ to 75℃ in 1 minute. What is the temperature of the body at the end of 3 minutes? (Take log(1.5)=0.4)

(a) 40℃

(b) 47.5℃

(c) 42.5℃

(d) 50℃

I have been asked this question in semester exam.

This interesting question is from Newton’s Law of Cooling and Escape Velocity in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right choice is (b) 47.5℃

Easiest explanation: By Newton’s law of cooling w.k.t T = t2 + (t1-t2) e^-kt, given t1=100℃, t2=25℃

when t=1 –> T(1) = 25 + 75 e^-k = 75℃ –> 50/75 = 2/3 = e^-k

–> 3/2=e^k taking log k=log(1.5)=0.4.

to find T when t=3 minute using the value of k we get

T = 25 + 75e^-0.4*3 = 47.5℃……e^-1.2=0.3.