ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 10.2 transmission of heat

A body cools in \(7\) minutes from \(60^{\circ}\,C\) to \(40^{\circ}\,C\). The temperature of the surrounding is \(10^{\circ}\,C\). The temperature of the body after the next \(7\) minutes will be

  1. A \(32^{\circ}\,C\)
  2. B \(30^{\circ}\,C\)
  3. C \(28^{\circ}\,C\)
  4. D \(34^{\circ}\,C\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(28^{\circ}\,C\)

Step-by-step Solution

Detailed explanation

using average rate of Newton's law of cooling \(\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_s\right)\) Given \(\frac{60-40}{7}=K(50-10) \ldots\) (i) \(\frac{40-T}{7}=K\left(\frac{40+T}{2}-10\right) \ldots\) (ii) From \((i)\) and \((ii)\) \(T =28^{\circ}\,C\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app