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# A cup of coffee is poured from a pot, whose contents are ${95^ \circ }$ C into a non-insulated cup in a room at ${20^ \circ }$ C. After a minute, the coffee has cooled to ${90^ \circ }$ C. How much time is required before the coffee reaches a drinkable temperature of ${65^ \circ }$ C ?A.4.3 minutesB.5.7 minutesC.6.0 minutesD.7.4 minutes

Last updated date: 17th Jun 2024
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Hint: In differential ​calculus we learned that the derivative of ln(x) is 1/x. Integration goes the other way: the integral (or antiderivative) of 1/x should be a function whose derivative is 1/x. As we just saw, this is ln(x).

Complete step by step solution:
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,
$\ln \dfrac{{45}}{{75}} = - 0.06899t \\ \Rightarrow t = 7.4\min utes \\$
Thus the required time is 7.4 minutes.

Note: In solving these types of questions never ignore the constant C. Always try to find out the value of the constant as most of the time we ignore it and get the wrong answer.