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JEE Mains · Physics · STD 11 - Transmission of heat
  • JEE Mains 2023
  • Medium
  • MCQ

A bowl filled with very hot soup cools from \(98^{\circ}\,C\) to \(86^{\circ}\,C\) in \(2\) minutes when the room temperature is \(22^{\circ}\,C\). \(..........\,minutes\) long it will take to cool from \(75^{\circ}\,C\) to \(69^{\circ}\,C\) ?

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Answer & Solution

Correct Answer

(B) \(1.4\)

Step-by-step Solution

Detailed explanation

\(\frac{\Delta Q }{\Delta t }=- K \left( T - T _0\right)\)

\(\frac{\Delta Q }{\Delta t }=- K \left( T _{ avg }- T _0\right)\)

\((i)\) \(\frac{ ms \times 12}{2}=- K \left(\frac{98+86}{2}-22\right)\)

\(6=-\frac{ K }{ ms }\left[\frac{98+86}{2}-22\right]\)

\(6=-\frac{ K }{ ms }[70]..........(i)\)

\((ii)\) \(\frac{ ms \times 6}{\Delta t }=- K \left(\frac{75+69}{2}-22\right)\) \(\frac{6}{\Delta t }=-\frac{ K }{ ms }(50) \ldots \ldots \ldots \ldots\) (ii)

\((ii)\) \(\div\) \((i)\)

\(\frac{6}{\Delta t (6)}=\frac{50}{70}\)

\(\Delta t =\frac{7}{5}=1.4\,min\)

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