Question

The temperature T of a cooling object drops at a rate proportional to the difference $\left( {T - S} \right)$, where S is the constant temperature of the surrounding medium. If initially $T = 150^\circ C$, find the temperature of the cooling object at any time ‘t’.

Answer 1

Hint: We are given the rate of change of temperature of a body. Using that we can form a differential equation by adding a positive proportionality constant. We can integrate the differential equation to get the required equation of temperature at any time. We can use the given initial condition to eliminate the constant of integration to get the required answer.

Complete step by step Answer:

We are given that the rate of cooling of a body with temperature T is proportional to $\left( {T - S} \right)$. We can write this in differential form.
$\dfrac{{dT}}{{dt}}\alpha - \left( {T - S} \right)$
We put the negative sign here as the temperature is decreasing.
Let $k > 0$ be the proportionality constant. Then we get the differential equation,
$ \Rightarrow \dfrac{{dT}}{{dt}} = - k\left( {T - S} \right)$.
Now we can solve the differential equation. For solving we use the method of separation of the variable.
$ \Rightarrow \dfrac{{dT}}{{\left( {T - S} \right)}} = - kdt$
Now we have same variables on same side of the equation. So, we can integrate,
$ \Rightarrow \int {\dfrac{{dT}}{{\left( {T - S} \right)}}} = \int { - kdt} $
We know that $\int {\dfrac{{dx}}{{x - a}} = \log \left( {x - a} \right)} $ and $\int {dx} = x$.
$ \Rightarrow \log \left( {T - S} \right) = - kt + C$ .. (1)
On applying the initial condition $T = 150^\circ C$ at $t = 0$, we get,
$ \Rightarrow \log \left( {150 - S} \right) = - k \times 0 + C$
$ \Rightarrow C = \log \left( {150 - S} \right)$ … (2)
On substituting equation (2) in (1), we get,
$ \Rightarrow \log \left( {T - S} \right) = - kt + \log \left( {150 - S} \right)$
On rearranging, we get, .
$ \Rightarrow \log \left( {T - S} \right) - \log \left( {150 - S} \right) = - kt$
We know that $\log a - \log b = \log \dfrac{a}{b}$. On applying this, we get,
$ \Rightarrow \log \left( {\dfrac{{T - S}}{{150 - S}}} \right) = - kt$
We can take the exponents on both sides of the equation.
$ \Rightarrow {e^{\log \left( {\dfrac{{T - S}}{{150 - S}}} \right)}} = {e^{ - kt}}$
We know that ${e^{\log a}} = a$, on applying this, we get,
$ \Rightarrow \dfrac{{T - S}}{{150 - S}} = {e^{ - kt}}$
On multiplying both sides of the equation with $150 - S$, we get,
$T - S = \left( {150 - S} \right){e^{ - kt}}$
On simplification, we get,
$T = S + \left( {150 - S} \right){e^{ - kt}}$

Therefore, the temperature of the cooling body at time t is given by $T = S + \left( {150 - S} \right){e^{ - kt}}$ where S is the temperature of surrounding medium and $k > 0$ is the constant of proportionality.

Note: The method of converting a word problem into a mathematical equation is known as mathematical modeling. We use the concept of differential equations to represent the rate of cooling. In this question, we are given a proportionality. We need to put a constant called proportionality constant. We took $k > 0$ because if $k = 0$ the rate of change will be zero and if $k < 0$, the rate of change will reverse. We solve the differential equation using the method of variable separation. In this method, we separate the variables to either side of the equation and integrate them separately. After integration, we must add the constant of integration and the value of this constant can be found using the initial conditions. Such problems in which initial values are given are known as initial value problems.