Question

The temperature difference between a hot body suspended inside a room (which is maintained at a constant temperature) and the room's temperature becomes half the original value in a time interval $t_0$. In how much time will the temperature difference between the body and the surrounding becomes $\frac{1}{3}$rd of the original value?

• A. $t_0{\ln }_{2}3$
• B. $t_0\ln 2$
• C. $t_0\ln 3$
• D. $t_0\ln \frac{3}{2}$

Answer 1

The correct option is A $t_0{\ln }_{2}3$

Let ${\theta }_0\to$ room temperature
${\theta }_1\to$ Initial temperature of body , $\theta \to$ temperature of body at time t.
From Newton's Law of cooling,
$-\frac{d\theta }{dt}=k(\theta -{\theta }_0)\Rightarrow \underset{{\theta }_1}{\overset{\theta }{\int }}\frac{d\theta }{\theta -{\theta }_0}=\underset{0}{\overset{t}{\int }}\text{-k dt}$
$\Rightarrow \ln [\theta -{\theta }_0]-\ln [{\theta }_1-{\theta }_0]=-\text{kt}\Rightarrow \frac{\theta -{\theta }_0}{{\theta }_1-{\theta }_0}=e^{-\text{kt}}$
$\Rightarrow \theta -{\theta }_0=\Delta \theta =\left(\Delta {\theta }_0\right)e^{-\text{kt}}\text{where}\Delta {\theta }_0={\theta }_1-{\theta }_0$
We have for first case, $\frac{1}{2}\Delta {\theta }_0=\Delta {\theta }_0{e}^{-{\text{kt}}_0}\Rightarrow e^{-k{t}_0}=\frac{1}{2}$
$\Rightarrow 2=e^{k{t}_0}\Rightarrow k{t}_0=\ln \left(2\right)....\left(1\right)$
Similarly in second case, $\frac{1}{3}\Delta {\theta }_0=\Delta {\theta }_0{e}^{-kt}\Rightarrow e^{kt}=3$
$\Rightarrow kt=\ln \left(3\right)....\left(2\right)$
$\frac{\left(2\right)}{\left(1\right)}\Rightarrow \frac{t}{t_0}=\frac{\ln \left(3\right)}{\ln \left(2\right)}\Rightarrow t=t_0{\ln }_{2}3$