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Question: A hot body is placed in the air and is cooled according to Newton’s law of cooling. The rate of decr...

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=0t = 0, ln⁡ln⁡(x)k\ln \dfrac{{\ln (x)}}{k} time, the body will lose half the maximum temperature it can lose. Find the value of xx.

Explanation

Solution

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:
dθdt=−k(θ−θ0)\dfrac{{d\theta }}{{dt}} = - k\left( {\theta - {\theta _0}} \right)
Here θ0{\theta _0} is the temperature of the surroundings and θ\theta is the temperature of the body at t=0.

Complete step by step solution:
According to the Newton’s law of cooling-
dθdt=−k(θ−θ0)\dfrac{{d\theta }}{{dt}} = - k\left( {\theta - {\theta _0}} \right) ------- (1)
Here θ0{\theta _0} is the temperature of surroundings and θ\theta is the temperature of the body at t=0.

Now, consider θ=θ1\theta = {\theta _1} at t=0 and modify the equation (1),
dθ(θ−θ0)=−kdt\dfrac{{d\theta }}{{\left( {\theta - {\theta _0}} \right)}} = - kdt
Now, integrate on both sides by applying the limits of t and θ\theta , we get-
∫θ1θdθ(θ−θ0)=−k∫0tdt\int\limits_{{\theta _1}}^\theta {\dfrac{{d\theta }}{{\left( {\theta - {\theta _0}} \right)}}} = - k\int\limits_0^t {dt}
⇒ln⁡(θ−θ0)(θ1−θ0)=−kt\Rightarrow\ln \dfrac{{\left( {\theta - {\theta _0}} \right)}}{{\left( {{\theta _1} - {\theta _0}} \right)}} = - kt
⇒(θ−θ0)=(θ1−θ0)e−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.
θ=θ0\theta = {\theta _0}
So, the total heat loss =ΔQm=ms(θ1−θ0)\Delta {Q_m} = ms\left( {{\theta _1} - {\theta _0}} \right). According to this question, body is losing half of its maximum value, which is- ΔQm2ms=(θ1−θ0)2\dfrac{{\Delta {Q_m}}}{{2ms}} = \dfrac{{\left( {{\theta _1} - {\theta _0}} \right)}}{2}

Let ΔQm2\dfrac{{\Delta {Q_m}}}{2} heat be lost in time t1{t_1}.
t1{t_1} =θ1−θ1−θ02{\theta _1} - \dfrac{{{\theta _1} - {\theta _0}}}{2}
⇒t1\Rightarrow{t_1} =θ1+θ02\dfrac{{{\theta _1} + {\theta _0}}}{2}
Now, substitute the value of θ in equation (2),
We get-
(θ1+θ02−θ0)=(θ1−θ0)e−kt1\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−kt1=12{e^{ - k{t_1}}} = \dfrac{1}{2}
−kt1=ln⁡12- k{t_1} = \ln \dfrac{1}{2}
∴t1=ln⁡2k\therefore{t_1} = \dfrac{{\ln 2}}{k}
So, this will be the answer.

Hence, the value of xx 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.