Question

A body with an initial temperature ${{\theta }_{1}}$ is allowed to cool in surrounding which is at a constant temperature of ${{\theta }_{0}}({{\theta }_{0}}<{{\theta }_{1}})$, assume that the Newton’s law of cooling is obeyed. Let k = constant. The temperature of the body after time t is the best expressed by
$\begin{align}
  & A.\text{ }\left( {{\theta }_{1}}-{{\theta }_{0}} \right){{e}^{kt}} \\
 & B.\text{ }\left( {{\theta }_{1}}-{{\theta }_{0}} \right)1n\left( kt \right) \\
 & C.\text{ }\left( {{\theta }_{0}}+\left( {{\theta }_{1}}-{{\theta }_{0}} \right){{e}^{-kt}} \right) \\
 & D.\text{ }{{\theta }_{1}}{{e}^{-kt}}-{{\theta }_{0}} \\
\end{align}$

Answer 1

Hint: In order to get the answer of this question we will use the formula that obeys Newton’s law of cooling and by integrating the expression up to limits given we can find required expression.

Formula used:
$\dfrac{d\theta }{dt}=-k(\theta -{{\theta }_{0}})$

Complete step by step solution:
$\Rightarrow $According to the Newton’s law of cooling
$\dfrac{d\theta }{dt}=-k(\theta -{{\theta }_{0}})$
Where, k = constant
$\theta $= body temperature

${{\theta }_{0}}$= surroundings temperature
$\Rightarrow $ Now rearranging the above equation as,
$\dfrac{1}{\left( \theta -{{\theta }_{0}} \right)}=-k\text{ }dt$
$\Rightarrow $ Now as given in a question,
$\Rightarrow $At time t = 0 the initial temperature of the body is ${{\theta }_{1}}$ and the time t = t temperature of the body is$\theta $.

$\Rightarrow $ Now applying integration with the above given limits we will get,
\[\begin{align}
  & {{\Rightarrow }\int_{\theta_1
}^{\theta }}\dfrac{d\theta }{\theta -{{\theta }_{0}}}=-k{\int_{0}^{t}}dt \\
 & \Rightarrow \left[ \ln \left( \theta -{{\theta }_{0}} \right) \right]_{{{\theta }_{1}}}^{\theta }=-k\left[ t \right]_{0}^{t} \\
 & \therefore \left[ \ln \left( \theta -{{\theta }_{0}} \right)-\ln \left( {{\theta }_{1}}-{{\theta }_{0}} \right) \right]=-kt \\
\end{align}\]

$\Rightarrow $Applying log rule
$\therefore \ln \left( \dfrac{\theta -{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}} \right)=-kt$

$\Rightarrow $ Now,
$\begin{align}
  & \Rightarrow \left( \dfrac{\theta -{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}} \right)={{e}^{-kt}} \\
 & \Rightarrow \theta -{{\theta }_{0}}={{\theta }_{1}}{{e}^{-kt}}-{{\theta }_{0}}{{e}^{-kt}} \\
 & \therefore \theta ={{\theta }_{0}}+\left( {{\theta }_{1}}-{{\theta }_{0}} \right){{e}^{-kt}} \\
\end{align}$

So from the above expression we can state that option (C) is the correct solution.

Additional information:
Newton’s law of cooling:
According to Newton’s law of cooling the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surrounding.

Limitation:
The loss of heat from the body should happen by radiation only.
The difference in temperature between body and surroundings should be small. $\to $ The surrounding temperature needs to remain constant.
Newton’s law of cooling is given by,
$\dfrac{dT}{dt}=k\left( {{T}_{t}}-{{T}_{s}} \right)$
Where,
$k$ Is the Positive constant that depends on the area and nature of the surface of the body under consideration
${{T}_{t}}$ Is the temperature at time t and
${{T}_{s}}$ Is the surrounding temperature

Note:
When we are giving limits as for integration don’t mistaken limits as ${{\theta }_{0}}$ to ${{\theta }_{1}}$or ${{\theta }_{0}}$to $\theta $ because ${{\theta }_{0}}$ is already given as constant temperature and ${{\theta }_{1}}$ as initial temperature.