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Physics Question on thermal properties of matter

A body takes 1010 minutes to cool from 60∘C60^{\circ}C to 50∘C50^{\circ}C. The temperature of surroundings is constant at 25∘C25^{\circ}C. Then, the temperature of the body after next 1010 minutes will be approximately :

A

47∘C47^{\circ} C

B

41∘C41^{\circ} C

C

45∘C45^{\circ} C

D

43∘C43^{\circ} C

Answer

43∘C43^{\circ} C

Explanation

Solution

Time taken to cool from 60∘C60^{\circ} C to 50∘C=1050^{\circ} C =10 minutes
Temperature of surroundings =25∘C=25^{\circ} C Temperature of body in next 10 minutes =T=T
Therefore, 60−5010min⁡=kB[60+502−25]⇒kB30=1       (1)\frac{60-50}{10 \min }=k_{ B }\left[\frac{60+50}{2}-25\right] \Rightarrow k_{ B } 30=1\,\,\,\,\,\,\,(1)
and 60−T20min⁡=kB[60+T2−25]=kB[60+T−502]       (2)\frac{60-T}{20 \min }=k_{ B }\left[\frac{60+T}{2}-25\right]=k_{ B }\left[\frac{60+T-50}{2}\right]\,\,\,\,\,\,\,(2)
Taking ratio of Eqs. (1) and (2), we get
2060−T=30kBkB(10+T2)\frac{20}{60-T}=\frac{30 k_{ B }}{k_{ B }\left(\frac{10+T}{2}\right)}
⇒2060−T=305+T/2\Rightarrow \frac{20}{60-T}=\frac{30}{5+T / 2}
⇒20(5+T2)=30(60−T)\Rightarrow 20\left(5+\frac{T}{2}\right)=30(60-T)
⇒100+T 10=1800−30 T\Rightarrow 100+T\, 10=1800-30 \,T
⇒1800−100=30 T+10 T\Rightarrow 1800-100=30\, T+10 \,T
⇒1700=40 T\Rightarrow 1700=40 \,T
⇒T=170040=42.5∘C−43∘C\Rightarrow T=\frac{1700}{40}=42.5^{\circ} C -43^{\circ} C