TIFR CSE 2014 | Part A | Question: 2

Question

A body at a temperature of $30$ Celsius is immersed into a heat bath at $0$ Celsius at time $t = 0$. The body starts cooling at a rate proportional to the temperature difference. Assuming that the heat bath does not change in temperature throughout the process, calculate the ratio of the time taken for the body to reach $1$ Celsius divided by the time taken for the body to reach $5$ Celsius.

• A. $\log 5$

• B. $\dfrac{\log 29}{\log 25}$

• C. $\large e^{5}$

• D. $1 + \log_{6} 5$

• E. None of the above

Answer 1

Let's say the Temperature of body at time $t$  be $T_{t} $, which means $T_{0} = 30.$

Now let the time at which temperature is $5$ be $t_1,$ which means  $T_{t_1} = 5$

and the time at which temperature is $1$ be $t_2$ , which means  $T_{t_2} = 1.$

Question Asks us to find $\dfrac{t_2}{t_1}$.

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Now,$\text{ Temperature decrease(D) at time} t\propto  \text{Body Temperature}  - \text{Bath Temperature}$ 

                            $D\propto T_{t}  \;\{\text{As Bath Temp. is 0 & Body Temp. is  Tt}\}$

                            $D = k\times T_{t}$ {where $k$ is proportionality constant}

now, it gives      $T_{t+1} = T_{t} - D = T_{t} - k\times T_{t}   = (1-k) T_{t}$ 

Now,$T_0 = 30$

         $T_1 = 30 (1-k)$

         $T_2 = 30(1-k)^{2}$

         $T_{t_1} = 30(1-k)^{t_1}$           &        $T_{t_2} = 30(1-k)^{t_2}$

$\Rightarrow 30(1-k)^{t_1} = 5$

$\Rightarrow  t_1\times \log (1-k)  = \log \left(\dfrac{5}{30}\right) = \log \left(\dfrac{1}{6}\right)$    

Similarly   $\Rightarrow t_{2}\times \log (1-k)  = \log \left(\dfrac{1}{30}\right)$  

$\dfrac{t_{2}}{ t_{1}} =\dfrac{\log \left(\dfrac{1}{30}\right)}{\log \left(\dfrac{1}{6}\right)}$

$=\dfrac{\log (30-1)}{\log (6-1)} =\dfrac{\log(30)} {\log (6)}=\log_{6} 30$

$=\log_{6} (6\times 5) =\log_{6} 6  +\log_{6} 5 =1 +\log_{6} 5$

which is OPTION (D)

Answer 2

Answer can be found out using below method as well.