A cup of coffee cools from 90^{circ} mathrm{C} to 80^{circ} mathrm{C} in mathrm{t} minutes, when roo

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10-2.Transmission of Heat

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A cup of coffee cools from $90^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ in $\mathrm{t}$ minutes, when the room temperature is $20^{\circ} \mathrm{C}$. The time taken by a similar cup of coffee to cool from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ at a room temperature same at $20^{\circ} \mathrm{C}$ is :

A

$\frac{13}{10} \mathrm{t}$

B

$\frac{13}{5} \mathrm{t}$

C

$\frac{10}{13} \mathrm{t}$

D

$\frac{5}{13} \mathrm{t}$

(NEET-2021)

Solution

$\frac{d T}{d t}=K\left(T_{a v}-T_{0}\right)$

$\frac{10}{t}=K(85-20)$

$\frac{20}{t}=K(70-20)$

$\frac{t^{\prime}}{2 t}=\frac{65}{50}$

$\frac{t^{\prime}}{t}=\frac{130}{50}$

$t^{\prime}=\frac{13}{5} t$

Standard 11

Physics

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