# Solve this following

Question:

In 5 minutes, a body cools from $75^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ at room temperature of $25^{\circ} \mathrm{C}$. The temperature of

Solution:

By newton's law of cooling (with approximation)

$\frac{\Delta \mathrm{T}}{\Delta \mathrm{t}}=-\mathrm{C}\left(\mathrm{T}_{\text {avg }}-\mathrm{T}_{\mathrm{s}}\right)$

$1^{\mathrm{st}} \frac{-10^{\circ} \mathrm{C}}{5 \min }=-\mathrm{C}\left(70^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$

$\Rightarrow \quad C=\frac{2}{45} \min ^{-1}$

$2^{\text {nd }} \frac{T-65}{5 \min }=-C\left(\frac{T+65}{2}-25\right)=-\left(\frac{2}{45}\right)\left(\frac{T+15}{2}\right)$

$\Rightarrow \quad 9(\mathrm{~T}-65)=-(\mathrm{T}+15)$

$\Rightarrow \quad 10 \mathrm{~T}=570$

$\Rightarrow \quad \mathrm{T}=57^{\circ} \mathrm{C}$

Alternate Solution:

Newton's law of cooling (without approximation)

$\mathrm{T}_{\mathrm{P}}-\mathrm{T}_{\mathrm{S}}=\left(\mathrm{T}_{\mathrm{i}}-\mathrm{T}_{\mathrm{S}}\right) e^{-\mathrm{Ct}}$

$1^{\text {st }} \quad 65-25=(75-25) \mathrm{e}^{-5 \mathrm{C}} \Rightarrow \mathrm{e}^{-5 \mathrm{C}}=\frac{4}{5}$

$2^{\text {nd }} \quad \mathrm{T}-25=(65-25) \mathrm{e}^{-5 \mathrm{C}}=40 \times \frac{4}{5}=32$

$\mathrm{~T}=57^{\circ} \mathrm{C}$