Newton's Law Of Cooling

Classical Mechanics Level 2

A thermometer reading 8 0 ∘ F 80^\circ F is taken outside.
Five minutes later the thermometer reads 6 0 ∘ F 60^\circ F .
After another 5 minutes it reads 5 0 ∘ F 50^\circ F .

What is the temperature outside ( ( in ∘ F ) ? ^\circ F)?

Assume that this process follows Newton's law of cooling.


The answer is 40.

Rishabh Jain by Rishabh Jain , 23, India

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3 solutions

Let the initial and final temperatures of the body be ( T 1 T_1 ) and ( T 2 T_2 ), time taken for the temperature to change be t t and temperature of the surroundings be T T .

By Newton's law of cooling we have,

( T 1 − T 2 ) / t = K ( ( T 1 + T 2 ) / 2 − T ) (T1-T2)/t = K((T1+T2)/2 - T) , where K K is a constant.

Substituting the temperatures in of the above two cases, we obtain two equations.

4 = K ( 70 − T ) 4 = K(70-T) and 2 = K ( 55 − T ) 2 = K(55-T) .

Solving both equations we get T = 40 T=40 . Therefore, the temperature of the surrounding is 40 degrees Fahrenheit .

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Vikash Singh
Mar 19, 2016

Just guessed the answer in one go :D

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Newton's Law of Cooling is T(final)=T(ambient)+(T(initial)-T(ambient)*e^(-kt) where k=constant and t= time elapsed

In this case,

60=T(ambient)+(80-T(ambient)*e^(-5k) ( equation 1)

50=T(ambient)+(80-T(ambient)*e^(-10k) ( equation2)

k= log2/5 and T ambient is 40

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