JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2022
- Hard
- Numerical
In an experment ot verify Newton's law of cooling, a graph is plotted between, the temperature difference \((\Delta T )\) of the water and surroundings and time as shown in figure. The initial temperature of water is taken as \(80^{\circ} \,C\). The value of \(t _{2}\) as mentioned in the graph will be...........

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Answer & Solution
Correct Answer
(B) \(16\)
Step-by-step Solution
Detailed explanation
\(T - T _{0}=\left( T _{ i }- T _{0}\right) e ^{-\frac{ Bt }{ ms }}\)
\(6 \lambda=\ln 1.5\)
\(40=60 \,e ^{-\lambda(6)} \Rightarrow 6 \lambda=\ln 1.5\)
\(20=60 \,e ^{-\lambda t _{2}} \Rightarrow t _{2} \lambda=\ln 3\)
\(\frac{ t _{2}}{6}=\frac{\ln 3}{\ln 1.5}\)
\(\therefore t _{2}=16.25 min\)
So \(\approx 16\)
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