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Physics Question on thermal properties of matter

A body initially at 80∘^{\circ}C cools to 64∘^{\circ}C in 5 min and to 52∘^{\circ}C in 10 min. The temperature of the surrounding is

A

26∘^{\circ}C

B

16∘^{\circ}C

C

36∘^{\circ}C

D

40∘^{\circ}C

Answer

16∘^{\circ}C

Explanation

Solution

According to Newton?s law of cooling θ1−θ2t=K[θ1+θ22−θ0]\frac{\theta_{1}-\theta_{2}}{t}=K\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right] In the first case, 80−645=K[80+642−θ0]\frac{80-64}{5}=K\left[\frac{80+64}{2}-\theta_{0}\right] or3.2= K [58−θ0]....(i)\quad3.2=\,K\,\left[58-\theta_{0}\right]\quad\quad\quad\quad\quad\quad....\left(i\right) In the second case, 64−525=K[64+522−θ0]\, \frac{64-52}{5}=K\left[\frac{64+52}{2}-\theta_{0}\right] or2.4=K[58−θ0]....(ii)\quad 2.4 = K \left[58 - \theta_{0}\right]\quad \quad \quad \quad \quad \quad ....\left(ii\right) Dividing (i) by (ii), we get 3.22.4=72−θ058−θ0\frac{3.2}{2.4}=\frac{72-\theta_{0}}{58-\theta_{0}} or 185.6−3.2θ0=172.8−2.4θ0  \quad185.6 - 3.2\theta_{0} = 172.8 - 2.4\theta _{0} \,\,or   θ0=16∘C\,\,\theta _{0} = 16^{\circ}C