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Physics Question on Newtons Law of Cooling

A copper ball is heated to 100∘C100{}^\circ C . It cools to 80∘C80{}^\circ C in 5 minutes and to 65∘C65{}^\circ C in 10 minutes. The temperature of the surrounding is

A

13∘C13{}^\circ C

B

15∘C15{}^\circ C

C

18∘C18{}^\circ C

D

20∘C20{}^\circ C

Answer

20∘C20{}^\circ C

Explanation

Solution

: Let the temperature of surrounding be θ∘C\theta {}^\circ C ∴\therefore In first case, θ1−θ2t=K(θ1+θ22−θ)\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-\theta \right) Or (100−80)5=K(100+802−θ)\frac{(100-80)}{5}=K\left( \frac{100+80}{2}-\theta \right) or 4=K(90−θ)4=K(90-\theta ) ............... (i) In second case, temperature cools from 80∘C80{}^\circ C to 65∘C65{}^\circ C in (10−5)=5(10-5)=5 minutes. ∴\therefore 80−655=K(80+652−θ)\frac{80-65}{5}=K\left( \frac{80+65}{2}-\theta \right) or 3=K(72.5−θ)3=K(72.5-\theta ) ............... (ii) or 43=90−θ72.5−θ\frac{4}{3}=\frac{90-\theta }{72.5-\theta } or 270−3θ=290−4θ270-3\theta =290-4\theta or θ=290−270  or θ=20∘C\theta =290-270~~or\text{ }\theta =20{}^\circ C .