Question

How do I find the value of k in Newton’s law of cooling?

Answer 1

Hint: In this question, we will first know about Newton's law of cooling and its expression. This will help us to solve the question directly. Further, we will use a problem, where we will find the value of k with the help of the expression, to understand better.

Formula used:
$T(t)=T_S+(T_0-T_S)e^{-kt}$

Complete step by step solution:
As we know, Newton’s law of cooling is given by the expression:
$T(t)=T_S+(T_0-T_S)e^{-kt}$
Here,
$T(t)$ is the temperature of an object or body at a given time t
$T_S$ is the surrounding temperature
$T_0$ is the given initial temperature of the object and the last one is k, which is the constant
So, we can say that the value of k, depends on the other values.

Now, we will take a problem and we will find the value of k, to understand better.

It is given that the average coffee temperature at a particular coffee shop is 75 degree Celsius. Jack purchases a coffee from the local coffee shop. After 10 minutes, the coffee has cooled to 67 degree Celsius. The given temperature outside the coffee shop is steady at 16 degree Celsius. Now, assuming the coffee follows Newton's Law of Cooling, find the value of the constant k.

As we already know the expression for the Newton’s law of cooling is given as:
$T(t)=T_S+(T_0-T_S)e^{-kt}$

Now, putting the given values in the above equation we get:
$67=16+(75-16)e^{-10k}$
$\Rightarrow 51=59e^{-10k}$
$\Rightarrow {\displaystyle \frac{51}{59}}=e^{-10k}$

Now, we will take ln on both sides of the equation and solve further:
$\ln \left({\displaystyle \frac{51}{59}}\right)=\ln \left(e^{-10k}\right)$
$\Rightarrow \ln \left({\displaystyle \frac{51}{59}}\right)=-10k$$\Rightarrow k=-{\displaystyle \frac{1}{10}}\ln \left({\displaystyle \frac{51}{59}}\right)$
$\therefore k\approx 0.01457$

Therefore, we get the value of k, and also we learnt how to find the value of k in Newton’s law of cooling.

Additional information:
As we know that Isaac Newton created a formula that helps us to calculate the temperature of an object when it loses heat. Now, this heat from the object moves from the object to its surrounding. Further, the rate of the temperature change is proportional to the obtained temperature difference between the object and its surroundings. This formula can be used to find the temperature at some given time i.e., t. The S.I unit of temperature is the Kelvin (K), but degrees Celsius are common.

Note:
We should remember that the S.I. unit of temperature is Kelvin which is represented by K. Also, this question can be easily solved, when we know the expression of Newton’s law of cooling at some given time t. This expression is valid only for objects that release heat to its surroundings.