In an experiment to verigy Newton’s law of cooling, a graph is plotted between, the temperature difference $\left(\Delta T\right)$  of the water and surrounding and time as shown in figure. The initial temperature of water is taken as $80°C.$  The value of t₂ as mentioned in the graph will be _______.

 $\frac{dT}{dt}=-k\left(T-T_0\right)$

$\frac{dT}{\left(T-T_0\right)}=-kdt$

$\Rightarrow k=-\frac{1}{6}\mathcal{l}n\left(\frac{2}{3}\right)$ - (1)

Now $\frac{dT}{dt}=-k\left(T-T_0\right)$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

<p>&nbsp;<math><mrow><mfrac><mrow><mi>d</mi><mi>T</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>=</mo><mo>−</mo><mi>k</mi><mrow><mo> (</mo><mrow><mi>T</mi><mo>−</mo><msub><mrow><mi>T</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mo>)</mo></mrow></mrow></math></p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753240827php8AZMb1.jpeg" alt="" width="271" height="194" loading="lazy"></picture></div></div><p><math><mrow><mfrac><mrow><mi>d</mi><mi>T</mi></mrow><mrow><mrow><mo> (</mo><mrow><mi>T</mi><mo>−</mo><msub><mrow><mi>T</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mo>)</mo></mrow></mrow></mfrac><mo>=</mo><mo>−</mo><mi>k</mi><mi>d</mi><mi>t</mi></mrow></math></p><p><math><mrow><mo>⇒</mo><mi>k</mi><mo>=</mo><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>6</mn></mrow></mfrac><mi mathvariant="script">l</mi><mi>n</mi><mrow><mo> (</mo><mrow><mfrac><mrow><mn>2</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow></math>&nbsp;- (1)</p><p>Now&nbsp;<math><mrow><mfrac><mrow><mi>d</mi><mi>T</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>=</mo><mo>−</mo><mi>k</mi><mrow><mo> (</mo><mrow><mi>T</mi><mo>−</mo><msub><mrow><mi>T</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mo>)</mo></mrow></mrow></math></p><p>t = 10.257</p><p>t₂ = 6 + 10.257 = 16.257 minutes</p><p>16.25 and or 16</p>

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