Question

A calorimeter of water equivalent $6g$ is having a water of mass $64g$ up to a particular volume. Another identical calorimeter has liquid of mass $50g$ and specific heat $0.6\dfrac{cal}{g{}^\circ C}$ up to equal level. If both of them cool in the similar surroundings through equivalent range of temperature and the time taken for the water to cool is \[140s\], then the time taken for the liquid to become cool will be:
\[\begin{align}
  & A.72s \\
 & B.140s \\
 & C.36s \\
 & D.120s \\
\end{align}\]

Answer 1

Hint: Newton's law of cooling states that for a small difference in temperature between a material and its surrounding, the rate of cooling of the material is found to be directly proportional to the temperature difference and the exposed surface area. This can be expressed in like,
\[\dfrac{dQ}{dt}\propto \left( q-{{q}_{s}} \right)\]
Where\[q\] and \[{{q}_{s}}\] are the temperature which is corresponding to the object and surroundings. This will be helpful in solving this question.

Formula used:
\[\dfrac{{{m}_{1}}{{S}_{1}}+{{m}_{2}}{{S}_{2}}}{{{t}_{1}}}=\dfrac{{{m}_{1}}{{S}_{1}}+{{m}_{3}}{{S}_{3}}}{{{t}_{2}}}\]
Where ${{m}_{1}},{{m}_{2}}$ and ${{m}_{3}}$ the masses, ${{S}_{1}},{{S}_{2}}$ and ${{S}_{3}}$ are the specific heat capacities.

Complete step by step answer:
According to the newton’s law of cooling,
\[\dfrac{dQ}{dt}\propto \left( q-{{q}_{s}} \right)\]
As we all know,
\[Q=\dfrac{mS\Delta T}{t}\]
Where $t$ is the time taken to cool down and $\Delta T$ is the temperature difference
Therefore we can write that,
\[\dfrac{{{m}_{1}}{{S}_{1}}+{{m}_{2}}{{S}_{2}}}{{{t}_{1}}}=\dfrac{{{m}_{1}}{{S}_{1}}+{{m}_{3}}{{S}_{3}}}{{{t}_{2}}}\]
Where ${{m}_{1}},{{m}_{2}}$ and ${{m}_{3}}$the masses, ${{S}_{1}},{{S}_{2}}$ and ${{S}_{3}}$ are the specific heat capacities.
Here it is to be noted that the water is having a specific heat of $1ca{{\lg }^{-1}}$. Both ${{S}_{1}},{{S}_{2}}$ are the specific heats of water, so that it will be unity. ${{S}_{1}},{{S}_{2}}$
${{S}_{1}}={{S}_{2}}=1$
And also
${{S}_{3}}=0.6$
Substituting these in equation will give,
\[\dfrac{6\times 1+64\times 1}{140}=\dfrac{6\times 1+50\times 0.6}{{{t}_{2}}}\]
Rearranging these values will give the time taken to cool down,
\[{{t}_{2}}=72s\]
Hence the correct answer is given as option A.

Note: Newton’s law of cooling is the basic concept that is used to solve this question. Newton's law of cooling states that the rate of heat loss from a material is directly proportional to the difference in the temperatures between the material and its surroundings. Newton's Law of Cooling is helpful in studies of water heating as it can tell us how fast the hot water in the pipes cools off.