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Question

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A.4.3 minutes

B.5.7 minutes

C.6.0 minutes

D.7.4 minutes

Answer

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According to the question,

Temperature of the contents of the pot = $ {95^ \circ } $ C

Temperature of non-insulated cup in a room at $ {20^ \circ } $ C

After a minute, the coffee has cooled to $ {90^ \circ } $ C

Therefore, we have to find the time required when the coffee reaches a drinkable temperature of $ {65^ \circ } $ C

So,

Let T be the temperature of soup = $ {95^ \circ } $ C

And $T_0$be the temperature of surrounding = $ {20^ \circ } $ C

So the difference in the temperature of soup and surrounding is D = T- $ T_0 $

$

\therefore D = T - T_0 \\

\Rightarrow D = {95^ \circ } - {25^ \circ } \\

\Rightarrow D = {75^ \circ }C \\

$

$\therefore $ By newton’s law we can write $

dD = kDdt \\

\therefore this \Rightarrow \dfrac{{dD}}{D} = kdt \\

$

Now integrating both sides

$

\int {\dfrac{{dD}}{D} = \int {kdt} } \\

\Rightarrow \ln D = kt + C............(1) \\

$

At t = 0 , T = 95

Therefore D =95 – 20 = 75

$ \ln D = kt + C $

$

\therefore \Rightarrow \ln (75) = k(0) + C \\

\therefore C = \ln (75) \\

$

Now, substituting C in (1) we get

$

\Rightarrow \ln D = kt + \ln (75) \\

\Rightarrow \ln D - \ln (75) = kt \\

\Rightarrow \ln \dfrac{D}{{75}} = kt \\

\\

$

At t = 1 minute, T = 90 so D = 70

$

\Rightarrow \ln (\dfrac{{70}}{{75}}) = k(1) \\

\therefore k = - 0.06899 \\

\Rightarrow \ln \dfrac{D}{{75}} = - 0.06899t.............(2) \\

\Rightarrow \dfrac{D}{{75}} = {e^{ - 0.06899t}} \\

\Rightarrow D = 75{e^{ - 0.06899t}} \\

$

Now, coffee is drinkable when T = 65C and D = 45C

Thus, substituting these values in (2)

We get,

$

\ln \dfrac{{45}}{{75}} = - 0.06899t \\

\Rightarrow t = 7.4\min utes \\

$