Question

A body takes $10$ minutes to cool from ${60}^{o}C$ to ${50}^{o}C$. The temperature of surroundings is constant at ${25}^{o}C$. Then, the temperature of the body after next $10$ minutes will be approximately

• A. ${43}^{o}C$
• B. ${47}^{o}C$
• C. ${41}^{o}C$
• D. ${45}^{o}C$

Answer 1

The correct option is A ${43}^{o}C$

$-\frac{dT}{dt}=k(T-T_o)$

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

$T=$ Temp of body

$T_o=$ Temp. of surrounding $={25}^{o}C$

$t=10min=10\left(60\right)=600sec$

${\int }_{50}^{60}\frac{dT}{(T-T_o)}=-k[t{]}_o^{600}$

$\frac{2.303}{k}{\log }_e[T-T_o{]}_{50}^{60}=-k[600]$

$\Rightarrow \frac{2.303}{k}{\log }_e\left[\frac{60-25}{50-25}\right]=-600$

$\Rightarrow \frac{2.303}{k}{\log }_e\left[\frac{30}{25}\right]=-600$ ...(1)

Now suppose for next to min, temp falls from 50 to T

$\frac{2.303}{k}{\log }_e[T-T_o{]}_T^{50}=-600$

$\frac{2.303}{k}{\log }_e\left[\frac{50-25}{T-25}\right]=-600$ ...(2)

from (1) & (2)

${\log }_e\left(\frac{35}{25}\right)={\log }_e\left(\frac{25}{T-25}\right)$

$\frac{35}{25}=\frac{25}{(T-25)}$

$7T-175=125$

$7T=125+175$

$=300$

$T=\frac{300}{7}\approx {43}^{o}C$