Question

A hot body is placed in the air and is cooled according to Newton’s law of cooling. The rate of decrease of the temperature being K times the temperature difference from the surroundings, starting from \[t = 0\], \[\ln \dfrac{{\ln (x)}}{k}\] time, the body will lose half the maximum temperature it can lose. Find the value of \[x\].

Answer 1

Hint: As we know, according to Newton’s law of cooling, the rate of heat loss of a body is directly proportional to the temperature difference between the body and its surroundings. As the temperature difference increases then the rate of heat loss of a body also increases.

Formula used:
\[\dfrac{{d\theta }}{{dt}} = - k\left( {\theta - {\theta _0}} \right)\]
Here \[{\theta _0}\] is the temperature of the surroundings and \[\theta \] is the temperature of the body at t=0.

Complete step by step solution:
According to the Newton’s law of cooling-
\[\dfrac{{d\theta }}{{dt}} = - k\left( {\theta - {\theta _0}} \right)\] ------- (1)
Here \[{\theta _0}\] is the temperature of surroundings and \[\theta \] is the temperature of the body at t=0.

Now, consider \[\theta = {\theta _1}\] at t=0 and modify the equation (1),
\[\dfrac{{d\theta }}{{\left( {\theta - {\theta _0}} \right)}} = - kdt\]
Now, integrate on both sides by applying the limits of t and \[\theta \], we get-
\[\int\limits_{{\theta _1}}^\theta {\dfrac{{d\theta }}{{\left( {\theta - {\theta _0}} \right)}}} = - k\int\limits_0^t {dt} \]
\[\Rightarrow\ln \dfrac{{\left( {\theta - {\theta _0}} \right)}}{{\left( {{\theta _1} - {\theta _0}} \right)}} = - kt\]
\[\Rightarrow\left( {\theta - {\theta _0}} \right) = \left( {{\theta _1} - {\theta _0}} \right){e^{ - kt}}\] ------- (2)
Our body will release heat until the temperature of the body is equal to its surroundings.
\[\theta = {\theta _0}\]
So, the total heat loss =\[\Delta {Q_m} = ms\left( {{\theta _1} - {\theta _0}} \right)\]. According to this question, body is losing half of its maximum value, which is- \[\dfrac{{\Delta {Q_m}}}{{2ms}} = \dfrac{{\left( {{\theta _1} - {\theta _0}} \right)}}{2}\]

Let \[\dfrac{{\Delta {Q_m}}}{2}\] heat be lost in time ${t_1}$.
${t_1}$ =\[{\theta _1} - \dfrac{{{\theta _1} - {\theta _0}}}{2}\]
$\Rightarrow{t_1}$ =\[\dfrac{{{\theta _1} + {\theta _0}}}{2}\]
Now, substitute the value of θ in equation (2),
We get-
\[\left( {\dfrac{{{\theta _1} + {\theta _0}}}{2} - {\theta _0}} \right) = \left( {{\theta _1} - {\theta _0}} \right){e^{ - k{t_1}}}\]
After solving this equation, \[{e^{ - k{t_1}}} = \dfrac{1}{2}\]
\[ - k{t_1} = \ln \dfrac{1}{2}\]
\[\therefore{t_1} = \dfrac{{\ln 2}}{k}\]
So, this will be the answer.

Hence, the value of $x$ is 2.

Note:Newton’s law of cooling is very useful for studying water heating. Also in this question, while integrating, remember to take upper and lower limits in proper manner. Greater the difference in temperature between the system and surrounding, more rapidly the heat is transferred.