A block of metal is heated to a temperature much higher than the room temperature and placed in an evacuated…
A block of metal is heated to a temperature much higher than the room temperature and placed in an evacuated cavity. The curve which correctly represents the rate of cooling ( $T$ is temperature of the block and $t$ is the time.)
Solution
Key Idea Newton's law of cooling is given by expression,
$
-\frac{d T}{d t}=k^{\prime}\left(T-T_0\right)
$
where, $k^{\prime}=\frac{k}{m s}$ and negative sign shows the rate of heat loss.
The given expression can be rearranged by integrating as,
$
\int \frac{d T}{T-T_0}=-k^{\prime} \int d t
$
$
\begin{array}{rlrl}
\log _e\left(T-T_0\right) & =-k^{\prime} t+\log _e A\left(\because \log _e A=\text { Constant }\right) \\
& -\frac{d T}{\left(T-T_0\right)} & =k^{\prime} d t \\
\Rightarrow & \ln \left(T-T_0\right) & =-k^{\prime} t \\
\Rightarrow & & T & =e^{-k^{\prime} t}+T_0 \text { at } t \rightarrow \infty \\
\Rightarrow & & T & =T_0 \text {, at } t=0 \Rightarrow T \rightarrow \infty
\end{array}
$
Hence, the graph as shown below, shows temperature of a body (done) varies exponentially with time from $T$ to $T_0\left(T_0 < T\right)$.
Thus, the correct option is (b)
