JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2021
- Medium
- Numerical
In \(5\, minutes,\) a body cools from \(75^{\circ} {C}\) to \(65^{\circ} {C}\) at room temperature of \(25^{\circ} {C}\). The temperature of body at the end of next \(5\, minutes\) is \(......\,{ }^{\circ} {C} .\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(57\)
Step-by-step Solution
Detailed explanation
By newton's law of cooling (with approximation)
\(\frac{\Delta {T}}{\Delta {t}}=-{C}\left({T}_{{avg}}-{T}_{{s}}\right)\)
\(1^{\text {st }}\, \frac{-10^{\circ} {C}}{5\, {min}}=-{C}\left(70^{\circ} {C}-25^{\circ} {C}\right)\)
\({C}=\frac{2}{45}\, {min}^{-1}\)
\(2^{\text {nd }} \,\frac{{T}-65}{5 {min}}=-{C}\left(\frac{{T}+65}{2}-25\right)=-\left(\frac{2}{45}\right)\left(\frac{{T}+15}{2}\right)\)
\(9({T}-65)=-({T}+15)\)
\(10 {T}=570\)
\({T}=57^{\circ} {C}\)
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