Question

A body cools from 50a°C to 40°C in 5 minutes in surrounding temperature 20°C find temperature of body in next 5 minutes​

Answer 1

GiveN :

• Surrounding Temperature $\sf{T_s\ =\ 20^{\circ} C}$
• Initial temperature $\sf{T_0\ =\ 50^{\circ} C}$
• Final Temperature $\sf{T\ =\ 40^{\circ} C}$

To FinD :

• Temperature changes in 5 mins

SolutioN :

Newton's Law of Cooling :

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

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

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

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