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Question: In a room where the temperature is \({30^ \circ }\,C\) , a body cools from \({61^ \circ }\,C\) to \(...

In a room where the temperature is 30∘ C{30^ \circ }\,C , a body cools from 61∘ C{61^ \circ }\,C to 59∘ C{59^ \circ }\,C in 44 minutes. The time taken by the body to cool from 51∘ C{51^ \circ }\,C to 49∘ C{49^ \circ }\,C will be
A. 4 min⁡4\,\min
B. 5 min⁡5\,\min
C. 6 min⁡6\,\min
D. 8 min⁡8\,\min

Explanation

Solution

Here we have to apply Newton's law of cooling. The law of cooling by Newton states that a body's temperature adjusts at a rate equal to the temperature differential between the body and its surroundings.

Complete step by step answer:
The Newton’s Law of cooling is also qualified to have the condition that the variation in temperature is minor and that the essence of the process for heat transfer remains the same. As such it is analogous to a hypothesis that the coefficient of heat transfer, mediating between heat losses and variations in temperature, is constant. In heat conduction, this condition is normally satisfied, as the thermal conductivity of most materials is only weakly dependent on temperature.
Convection cooling is often said to be regulated by "Newton's law of cooling." When the coefficient of heat flow is independent of the temperature differential between material and atmosphere or reasonably independent, Newton's law is followed. For forced air and pumped liquid cooling, the rule holds well where the fluid velocity does not increase with the temperature difference rising.
The mathematical form of Newton’s law of cooling is given by:
dTdt=−bA(T−T∘) ⇒dTT−T∘=−bAdt  \dfrac{{dT}}{{dt}} = - bA\left( {T - {T_ \circ }} \right) \\\ \Rightarrow\dfrac{{dT}}{{T - {T_ \circ }}} = - bAdt \\\
On integrating we get:
1tln⁡(T2−T∘T1−T∘)=constant\dfrac{1}{t}\ln \left( {\dfrac{{{T_2} - {T_ \circ }}}{{{T_1} - {T_ \circ }}}} \right) = {\text{constant}} ....... (i)
Given,
When time =4s = 4s
T2=59∘ C ⇒T1=61∘ C ⇒T∘=30∘ C  {T_2} = {59^ \circ }\,C \\\ \Rightarrow{T_1} = {61^ \circ }\,C \\\ \Rightarrow{T_ \circ } = {30^ \circ }\,C \\\
When time =t s = t\,s
T2=49∘ C ⇒T1=51∘ C ⇒T∘=30∘ C  {T_2} = {49^ \circ }\,C \\\ \Rightarrow{T_1} = {51^ \circ }\,C \\\ \Rightarrow{T_ \circ } = {30^ \circ }\,C \\\
From equation (i) we get:
1tln⁡(T2−T∘T1−T∘)=constant ⇒14ln⁡(59−3061−30)=1tln⁡(49−3051−30) ⇒14ln⁡(2931)=1tln⁡(1921) ∴t=6 min⁡  \dfrac{1}{t}\ln \left( {\dfrac{{{T_2} - {T_ \circ }}}{{{T_1} - {T_ \circ }}}} \right) = {\text{constant}} \\\ \Rightarrow \dfrac{1}{4}\ln \left( {\dfrac{{59 - 30}}{{61 - 30}}} \right) = \dfrac{1}{t}\ln \left( {\dfrac{{49 - 30}}{{51 - 30}}} \right) \\\ \Rightarrow \dfrac{1}{4}\ln \left( {\dfrac{{29}}{{31}}} \right) = \dfrac{1}{t}\ln \left( {\dfrac{{19}}{{21}}} \right) \\\ \therefore t = 6\,\min \\\
Thus, the time taken by the body to cool from 51∘ C{51^ \circ }\,C to 49∘ C{49^ \circ }\,C will be 6 min⁡6\,\min .

Hence, option C is the correct answer.

Note: Here we have to see which temperature value should be put where. If we mistakenly put the wrong temperature value in an equation then the answer would be wrong.The cooling rate also depends on: (i) the area of the body's skin. (ii) The essence of the body's crust. (iii) Substance of the body surface (material effects of body wall conductivity).