Question

A body initially at $60\;{}^ \circ C$ cools to $50\;{}^ \circ C$ in$10\,\,\min $. What will be its temperature at the end of the next $10\,\,\min $. If the room temperature is $20\;{}^ \circ C$, Assume Newton’s law of cooling.
$A)\,\,42.50{}^ \circ C$
$B)\,\,45\;{}^ \circ C$
$C)\,\,40.46\;{}^ \circ C$
$D)\,\,44.23\;{}^ \circ C$

Answer 1

Hint: In the question it is given to use Newton’s law of cooling which is related to heat transfer between bodies, where the mode of heat transfer is through thermal radiation. The law states that the rate of transfer of heat between two bodies is directly proportional to temperature difference between the two bodies and depends exponentially on time.

Formula used: $T = {T_s} + \left( {{T_0} - {T_s}} \right){e^{ - Kt}}$
$T$ is temperature at any time $t$
${T_s}$ is temperature of surrounding
${T_0}$ is initial temperature
$K$ is cooling constant

Complete step by step answer:
It is given that initial temperature is $60\;{}^ \circ C$, and in $10\,\,\min $ it changes to $50\;{}^ \circ C$
Than applying Newton’s law of cooling
$
  50 = 20 + \left( {60 - 20} \right){e^{ - K10}} \\
  \dfrac{3}{4} = {e^{ - K10}} \\
$
From the above equation we get the value of constant part which is not given in the question, now again applying Newton’s law of cooling, but in this case what our final temperature was in first case it will become initial temperature in this case, also the time taken for this temperature change is same as was in previous case that is $10\,\,\min $.
Let the final temperature after $10\,\,\min $ be $T$.
$T = 20 + \left( {50 - 20} \right){e^{ - K10}}$
On substituting the value of ${e^{ - K10}}$ we get,
$
  T = 20 + 30 \times \dfrac{3}{4} \\
  T = 20 + \dfrac{{45}}{2} \\
  T = 42.50\;{}^ \circ C \\
$
So after $10\,\,\min $ the temperature of the body will be $42.50\;{}^ \circ C$.
So, the correct answer is “Option A”.

Additional Information:
Newton’s law of cooling is used to study heat transfer from one body to another body by radiation. Radiation is one of the three ways of heat transfer which are conduction, convection and radiation, radiation is a process of heat transfer in which energy is transferred by electromagnetic waves, the electromagnetic waves are sometimes visible to human eyes and sometimes not, depending on their frequency.

Note: In this question the time taken for temperature change is same in both cases that is from $60\;{}^ \circ C$ to $50\;{}^ \circ C$ and from $50\;{}^ \circ C$ to $42.50\;{}^ \circ C$ it took $10\,\,\min $, therefore we directly substituted for ${e^{ - K10}}$, if the time were different we would have substituted for $K$ that is the cooling constant.