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JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2018 · 15 April · Evening Shift · Online · Physics · Thermal Properties of Matter

A body takes 10 minutes to cool from 60^ C to 50^ C . The...

A body takes 10 minutes to cool from 60∘C60^{\circ} \mathrm{C} to 50∘C50^{\circ} \mathrm{C}. The temperature of surroundings is constant at 25∘C25^{\circ} \mathrm{C}. Then, the temperature of the body after next 10 minutes will be approximately

Options

  1. A43∘C43^{\circ} \mathrm{C}
  2. B47∘C47^{\circ} \mathrm{C}
  3. C41∘C41^{\circ} \mathrm{C}
  4. D45∘C45^{\circ} \mathrm{C}

Correct answer

A. 43∘C43^{\circ} \mathrm{C}

Step-by-step solution

According to Newton's law of cooling, (θ1−θ2t)=K(θ1+θ22−θ0)(60−5010)=K(60+502−25) and, (50−θ10)=K(50+θ2−25)\begin{aligned}&\left(\frac{\theta_1-\theta_2}{t}\right)=K\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right) \\ &\left(\frac{60-50}{10}\right)=K\left(\frac{60+50}{2}-25\right) \\ &\text{ and, }\left(\frac{50-\theta}{10}\right)=K\left(\frac{50+\theta}{2}-25\right) \end{aligned}Dividing eq. (i) by (ii), 10(50−θ)=60θ⇒θ=\frac{10}{(50-\theta)}=\frac{60}{\theta} \Rightarrow \theta=42.85^{{}^{\circ}} C≅\mathrm{C} \cong43^{{}^{\circ}} \mathrm{C}
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