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Question: A body cools from \(80^\circ C\) to \(70^\circ C\) in 5 minutes and to \(62^\circ C\) in the next 5 ...

A body cools from 80∘C80^\circ C to 70∘C70^\circ C in 5 minutes and to 62∘C62^\circ C in the next 5 minutes. Calculate temperature of the surrounding.

Explanation

Solution

Newton’s law of cooling says that the rate of change of the temperature with respect to time is equal to the difference of the temperature of an object and the temperature of the surrounding. The change in temperature decreases as the difference in temperature of the surrounding and the object decreases.

Formula used: The formula of change in temperature with time is given by,
⇒dTdt=−k(T−To)\Rightarrow \dfrac{{dT}}{{dt}} = - k\left( {T - {T_o}} \right)
Where average temperature is T, the time is t, the constant is k and the temperature of the surrounding isTo{T_o}.

Complete step by step solution:
It is given in the problem that a body cools from 80∘C80^\circ C to 70∘C70^\circ C in 5 minutes and to 62∘C62^\circ C in the next 5 minutes and we need to calculate the temperature of the surrounding.
For first decrease in temperature. The average temperature is equal to,
⇒T=80+702\Rightarrow T = \dfrac{{80 + 70}}{2}
⇒T=1502\Rightarrow T = \dfrac{{150}}{2}
⇒T=75∘C\Rightarrow T = 75^\circ C
The formula of change in temperature with time is given by,
⇒dTdt=−k(T−To)\Rightarrow \dfrac{{dT}}{{dt}} = - k\left( {T - {T_o}} \right)
Where average temperature is T, the time is t, the constant is k and the temperature of the surrounding isTo{T_o}.
The body cools from 80∘C80^\circ C to 70∘C70^\circ C in 5 minutes. According to Newton’s law of cooling.
⇒80−705=−k(75−To)\Rightarrow \dfrac{{80 - 70}}{5} = - k\left( {75 - {T_o}} \right)
⇒105=−k(75−To)\Rightarrow \dfrac{{10}}{5} = - k\left( {75 - {T_o}} \right)
⇒2=−k(75−To)\Rightarrow 2 = - k\left( {75 - {T_o}} \right)………eq. (1)
For the decrease temperature from 70∘C70^\circ C to 62∘C62^\circ C in 5 minutes.
The average temperature is equal to,
⇒T=70+622\Rightarrow T = \dfrac{{70 + 62}}{2}
⇒T=66∘C\Rightarrow T = 66^\circ C
The body cools from 70∘C70^\circ C to 62∘C62^\circ C in 5 minutes. According to Newton’s law of cooling.
⇒dTdt=−k(T−To)\Rightarrow \dfrac{{dT}}{{dt}} = - k\left( {T - {T_o}} \right)
⇒70−625=−k(66−To)\Rightarrow \dfrac{{70 - 62}}{5} = - k\left( {66 - {T_o}} \right)
⇒85=−k(66−To)\Rightarrow \dfrac{8}{5} = - k\left( {66 - {T_o}} \right)………eq. (2)
Dividing the equation (2) with equation (1) we get,
⇒810=66−To75−To\Rightarrow \dfrac{8}{{10}} = \dfrac{{66 - {T_o}}}{{75 - {T_o}}}
⇒8(75−To)=10(66−To)\Rightarrow 8\left( {75 - {T_o}} \right) = 10\left( {66 - {T_o}} \right)
⇒600−8To=660−10To\Rightarrow 600 - 8{T_o} = 660 - 10{T_o}
⇒2To=60\Rightarrow 2{T_o} = 60
⇒To=30∘C\Rightarrow {T_o} = 30^\circ C.

The surrounding temperature is equal toTo=30∘C{T_o} = 30^\circ C.

Note: It is advisable for students to understand and remember the formula of Newton's law of cooling as it is very helpful in solving problems like these. The average temperature is taken of the initial and final temperature of the object.