Question

The temperature of a body falls from 40℃ to 36℃ in 5 minutes when placed in a surrounding of constant temperature 16℃. Find the time taken for the temperature of the body to become 32℃.

Answer 1

Answer:

=6.1min . Alternative method The mean temperature of the body as it cools from 40∘C to 36∘C is 40∘C+36∘C2=38∘C . The rate of decrease of temperature is 40∘C+36∘C5min=0.80∘C min^(-1).

Answer 2

Step-by-step explanation:

GiveN :

Surrounding Temperature $\sf{T_s\ =\ 20^{\circ} C}$

Initial temperature $\sf{T_0\ =\ 50^{\circ} C}$

Final Temperature $\sf{T\ =\ 40^{\circ} C}T = 40$

To FinD :

Temperature changes in 5 mins

• SolutioN :

Newton's Law of Cooling :

$\begin{gathered}\implies \boxed{\boxed{\sf{T\ =\ T_s\ +\ (T_0\ -\ T_s)e^{-kt}}}} \\ \\ \\ \\ \implies \sf{\dfrac{T\ -\ T_s}{T_0\ -\ T_s}\ =\ e^{-kt}}\end{gathered}$

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$\sf{Where} \: \begin{cases} \sf{ e^{-kt}\ is\ constant} \end{cases}$

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$\begin{gathered}\implies \sf{\dfrac{40\ -\ 20}{50\ -\ 20}\ =\ e^{-kt}} \\ \\ \\ \\ \implies \sf{\dfrac{20}{30}\ =\ e^{-kt}} \\ \\ \\ \\ \implies {\underline{\boxed{\sf{e^{-kt}\ =\ \dfrac{2}{3}}}}}\end{gathered}$

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________________

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• As constant will not be changed, So constant in first 5mins will be equal to constant in next 5mins. So, after 5mins :

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$\begin{gathered}\sf{Given} \begin{cases} \sf{Initial\ Temperature\ (T_0)\ =\ 40^{\circ}} \\ \\ \sf{Final\ Temperature\ =\ T} \end{cases} \\ \\ \\ \\ \implies \sf{\dfrac{T\ -\ T_s}{T_0\ -\ T_s}\ =\ e^{-kt}} \\ \\ \\ \\ \implies \sf{\dfrac{T\ -\ 20}{40\ -\ 20}\ =\ \dfrac{2}{3}} \\ \\ \\ \\ \implies \sf{\dfrac{T\ -\ 20}{20}\ =\ \dfrac{2}{3}} \\ \\ \\ \\ \implies \sf{T\ -\ 20\ =\ \dfrac{2\ \times\ 20}{3}} \\ \\ \\ \\ \implies \sf{T\ -\ 20\ =\ \dfrac{40}{3}} \\ \\ \\ \\ \implies \sf{T\ =\ \dfrac{40}{3}\ +\ 20} \\ \\ \\ \\ \implies \sf{T\ =\ \dfrac{40\ +\ 3(20)}{3}} \\ \\ \\ \\ \implies \sf{T\ =\ \dfrac{40\ +\ 60}{3}} \\ \\ \\ \\ \implies \underline{\boxed{\sf{T\ =\ \dfrac{100}{3}^{\circ}}}}\end{gathered}$