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Physics Question on thermal properties of matter

Hot water cools from 60∘C60^{\circ}C to 50∘C50^{\circ}C in the first 1010 minutes and to 42∘C42^{\circ}C in the next 1010 minutes. Then the temperature of the surroundings is

A

15∘C15^{\circ}C

B

10∘C10^{\circ}C

C

20∘C20^{\circ}C

D

30∘C30^{\circ}C

Answer

10∘C10^{\circ}C

Explanation

Solution

According to Newton's law of cooling
θ2−θ1t=K[θ1+θ22−θs]\frac{\theta_{2}-\theta_{1}}{t}=K\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{s}\right]
where, θs\theta_{s} is the temperature of the surroundings.
60−5010=K[60+502−θs]\frac{60-50}{10}=K\left[\frac{60+50}{2}-\theta_{s}\right]
1=K[55−θs]1=K\left[55-\theta_{s}\right]...(i)
Similarly, 50−4210=K(46−θs)\frac{50-42}{10}=K\left(46-\theta_{s}\right)
810=K(46−θs)\frac{8}{10}=K\left(46-\theta_{s}\right)...(ii)
Dividing E (i) by E (ii), we get
108=K(55−θs)K(46−θs)\frac{10}{8}=\frac{K\left(55-\theta_{s}\right)}{K\left(46-\theta_{s}\right)}
⇒θs=10∘C\Rightarrow \theta_{s}=10^{\circ} C