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