Question

A bottle of mineral water at a room temperature of 72℉ is kept in a refrigerator where the temperature is 44℉.After half an hour water cooled to 61℉.What is the temperature of the body in another half an hour?(Take log  \(\frac{28}{17}\) = 0.498, e^-0.99=0.37)

(a) 18℉

(b) 9.4℉

(c) 54.4℉

(d) 36.4℉

I had been asked this question in unit test.

I need to ask this question from Newton’s Law of Cooling and Escape Velocity in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct choice is (c) 54.4℉

The explanation is: By Newton’s law of cooling w.k.t T=t2+(t1-t2) e^-kt, given t1=72℉, t2=44℉

At t=half an hour = 30mts T=61℉, finding k using the given values i.e

61=44+28e^-k30 –> \(\frac{17}{28}\) = e^-k30 or \(\frac{28}{17}\) = e^k30 taking log, log  \(\frac{28}{17}\) = 30k –> k=0.0166

to find T when t = 30mts + 30mts = 60mts

T = 44 + 28e^-(0.0166)30 = 54.4℉.