Question

If a liquid takes 30s in cooling from \[{95^0}C\] and to \[{90^0}C\] and 70s in cooling from \[{55^0}C\] to \[{50^0}C\] then temperature of room is:
(A) \[{16.5^0}C\]
(B) \[{22.5^0}C\]
(C) \[{28.5^0}C\]
(D) \[{32.5^0}C\]

Answer 1

Hint Temperature change w.r.t time takes first-order kinetics. Which means, the rate of change of the temperature is proportional to the difference in the temperature with the constant of proportionality being K. The room temperature is the same in both cases. Hence the difference in temperature is known in both cases in terms of the room temperature, say \[{T_{room}}\]. The rates in the two cases can be related using this temperature difference and the constant of proportionality K which is common in both cases. So, there are two equations and two variables (K and T)

Complete step-by-step solution For cooling from initial temperature \[{T_1}\] to final temperature \[{T_2}\], the relation is given by,
$\dfrac{{{T_1} - {T_2}}}{t} = K\left( {\dfrac{{{T_1} + {T_2}}}{2} - {T_0}} \right)$
Where, K is a constant, t is the time taken for cooling and \[{T_0}\] is the room temperature.
Case (1): Cooling from \[{95^0}C\] to \[{90^0}C\] in 30s
$\begin{gathered}
  \dfrac{{95 - 90}}{{30}} = K\left( {\dfrac{{95 + 90}}{2} - {T_0}} \right) \\
  \dfrac{1}{6} = K\left( {\dfrac{{185}}{2} - {T_0}} \right) \\
  \dfrac{1}{6} = K\left( {92.5 - {T_0}} \right) \Rightarrow (1) \\
\end{gathered} $
Case (2): Cooling from \[{55^0}C\] to \[{50^0}C\] in 70s
$\begin{gathered}
  \dfrac{{55 - 50}}{{70}} = K\left( {\dfrac{{55 + 50}}{2} - {T_0}} \right) \\
  \dfrac{1}{{14}} = K\left( {\dfrac{{105}}{2} - {T_0}} \right) \\
  \dfrac{1}{{14}} = K\left( {52.5 - {T_0}} \right) \Rightarrow (2) \\
\end{gathered} $
Dividing equation (1) by (2), we get
$\begin{gathered}
  \dfrac{7}{3} = \dfrac{{92.5 - {T_0}}}{{52.5 - {T_0}}} \\
  367.5 - 7{T_0} = 277.5 - 3{T_0} \\
  4{T_0} = 90 \\
  {T_0} = 22.5{}^0C \\
\end{gathered} $

Hence the room temperature is \[{22.5^0}C\] and the correct option is B.

Note The above formula is derived from the rate of cooling expression. When a body cools by radiation the rate depends on surface, area, mass of the body, specific heat of the body, temperature of the body and the surroundings.
There can be another question in the same scenario asking for the value of K. To answer that, one only needs to equate the two temperatures. This will leave only one equation with one variable K.