Question

A body coals in \[T\] minutes from $60{}^\circ C$ to $40{}^\circ C$. If the temperature of the surroundings is $10{}^\circ C,$ the temperature after next $7$ minutes will be.
(a) $32{}^\circ C$
(b) $38{}^\circ C$
(c) $22{}^\circ C$
(d) None of these

Answer 1

Hint: Newton’s law cooling states that the rate of heat loss of a body is directly proportional to the difference in temperature between body and surrounding.
Formula used:-
Newton’s law of cooling
$\dfrac{dT}{dt}=K\left( T-{{T}_{surrounding}} \right)$
On integrating we get,
$\int{\dfrac{dT}{dt}=\int{K\left( T-{{T}_{surrounding}} \right)}}$
We get,
${{T}_{final}}-{{T}_{surrounding}}={{e}^{-Kt}}$
${{T}_{initial}}-{{T}_{surrounding}}$

Complete Step by Step Answer:
By Newton’s law of cooling we have
${{T}_{final}}-{{T}_{surrounding}}={{e}^{-Kt}}$
${{T}_{initial}}-{{T}_{surrounding}}$
Where,
${{T}_{initial}}=60{}^\circ C$
Time taken $\left( t \right)=7\min .$
$\dfrac{40-10}{60-10}={{e}^{7K}}...(i)$
According to the given data we have to calculate final temperature after $7\min .$ Now initial temperature becomes $40{}^\circ C$ and time taken $=7\min .$
$\therefore \dfrac{{{T}_{f}}-10}{40-10}={{e}^{-7K}}...(ii)$
Equation $(i)\And (ii)$
$\dfrac{40-10}{60-10}=\dfrac{{{T}_{f}}-10}{40-10}$
${{T}_{f}}=10+18=28{}^\circ C$

$\therefore $ Option (d) i.e. none of these is correct

Additional information: Rate of cooling is faster at start and decreases or slows down as the difference of temperature goes on decreasing.

Note:
Newton’s law of cooling is applicable when the temperature difference between the object and its surrounding is small compared to temperature of the object.