Question

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)

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Newton's law of cooling is given by expression,
$-\frac{dT}{dt}=k^{\prime }\left(T-T_0\right)$
where, $k^{\prime }=\frac{k}{ms}$ and negative sign shows the rate of heat loss.
The given expression can be rearranged by integrating as,
$\int \frac{dT}{T-T_0}=-k^{\prime }\int dt$
${\log }_e\left(T-T_0\right)=-k^{\prime }t+{\log }_{e}A$
$(\because {\log }_{e}A=$ Constant
$-\frac{dT}{\left(T-T_0\right)}=k^{\prime }dt$
$\Rightarrow \ln \left(T-T_0\right)=-k^{\prime }t$
$\Rightarrow T=e^{-k^{\prime }t}+T_0$ at $t\to \infty$
$\Rightarrow T=T_0,$ at $t=0\Rightarrow T\to \infty$
Hence, the graph as shown below,
shows temperature of a body (done) varies exponentially with time from
$T$ to $T_0\left(T_0<T\right)$.