Question

Water at ${100}^{o}C$ cool in 10 minutes to ${88}^{o}C$, in the room temperature of ${25}^{o}C$ find the temperature of water after 20 minutes.

Answer 1

From Newton's law of coding

$\frac{-d\theta }{dt}\alpha (\theta -{\theta }_s)$

where $\frac{-d\theta }{dt}=$ rate of coding

$\theta =$ averagetemp of body

${\theta }_s=$ averagetemp surrounding

Here, $\frac{12}{10}=k(\frac{88+100}{2}-25)$

$\Rightarrow 1.2=k(94-25)$

$\Rightarrow 1.2=k\left(69\right)----\left(1\right)$

and $\frac{100-\theta }{20}=k(\frac{100+\theta }{2}-25)$

$\Rightarrow 100-\theta =k(1000+10\theta -500)----\left(2\right)$

equation $\left(2\right)\Rightarrow \frac{1.2}{100-\theta }=\frac{69}{500+10\theta }$

$\Rightarrow 60\theta +12\theta =6900-690$

$\Rightarrow 81\theta =6300$

$\Rightarrow \theta ={77.78}^{o}C$