Question

A thermometer at room temperature 30°C is dipped suddenly into a bath of boiling water at 100°C. It takes 30 seconds to reach 96.5°C. The time required to reach a temperature of 98°C is

• A. 32.5 s
• B. 34.6 s
• C. 35.6 s
• D. 38.6 s

Answer 1

The correct option is C 35.6 s

Thermometer formula, considering the initial conditions,

$\theta \left(t\right)={\theta }_f(1-e^{-t/\tau })+{\theta }_i{e}^{-t/\tau }$

where,
${\theta }_f=finaltemperature\left(°C\right)$

$\theta \left(t\right)=temperatureattime\left(t\right)$

${\theta }_i=initialtemperature\left(°C\right)$

$96.5=100(1-e^{-30/\tau })+30e^{-30/\tau }$

$=100-100e^{-30/\tau }+30e^{-30/\tau }$

$=100-70e^{-30/\tau }$

$70e^{-30/\tau }=3.5$

$e^{-30/\tau }=\frac{3.5}{70}=0.05$

Taking log both sides we get,

$log{e}^{-30/\tau }=log0.05$

$-\frac{30}{\tau }=-2.99=-3.00$

$\therefore t=\frac{30}{3}=10sec.$

$for\theta \left(t\right)={98}^{\circ }C$

$98=100(1-e^{-t/10})+30e^{-t/10}$

$98=100-70e^{-t/10}$

$70e^{-t/10}=2$

$\therefore e^{-t/10}=2/70$

Taking log both sides we get,

$-\frac{t}{10}=log\frac{2}{70}=-35.55$

$-t=-3.555$

$t=35.55=35.6sec$