Question

A body cools from 91 to 89 in 5 minutes in 20 .time for cooling from 71 to 69
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Answer 1

Answer:

A body cools from 91℃ to 89℃ in 5 minutes when room temperature is 20℃. then it will cool from 71℃ to 69℃ in. ... ATQ time is 5 minutes

Explanation:

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Answer 2

Given:

Time taken for cooling from $91^{o}$ to $89^{o}$ $=5min$

To find:

Time taken for cooling from $71^{o}$ to $69^{o}$.

Solution:

The Newton's Law of cooling states that the rate of cooling of a hot substance from a higher temperature to a comparatively lower temperature is directly proportional to the difference in the temperatures.

                                  $\frac{dQ}{dt}$ ∝ $(T_{2}- T_{1} )$

                                $\frac{dQ}{dt}=K(T_{2}- T_{1} )$

Where, $T_{1}$ is the initial temperature and $T_{2}$ is the temperature at time t.

According to the question,

$\frac{dQ}{5}=K(89-91)$

$\frac{dQ}{K}=5(-2)$

$\frac{dQ}{K}= -10$

Now,

Time taken for cooling from $71^{o}$ to $69^{o}$, we get,

$\frac{dQ}{dt}=K(69-71 )$

$\frac{dQ}{dt}=K(-2)$

$\frac{dQ}{K(-2)}=dt$

We know, $\frac{dQ}{K}= -10$

$t=\frac{-10}{-2}$

$t=5min$

Final answer:

Hence, the time taken to cool from 71° to 69° is 5 min.