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Question: A cup of a coffee at temperature \({100^0}\) C is placed in a room whose temperature is \({15^0}\)C ...

A cup of a coffee at temperature 1000{100^0} C is placed in a room whose temperature is 150{15^0}C and it cools to 600{60^0}C in 5 minutes. Find its temperature after a further interval of 5 minutes.

Explanation

Solution

Hint: Here, we need to find the temperature after a further interval of 5 minutes by using Newton's law of cooling.

Complete step-by-step answer:
Given, the initial temperature of the coffee T0=1000 C{T_0} = {100^0}{\text{ C}}
The temperature of the room is TiT_i = 150{15^0}C.
Let ‘T’ be the temperature at time ‘t’ minutes.
By Newton's law of cooling;
dTdtα(T−Ti) ⇒dTdtα(T−15) ∴dTdt=−k(T−15)  \dfrac{{dT}}{{dt}}\alpha (T - T_i) \\\ \Rightarrow \dfrac{{dT}}{{dt}}\alpha (T - 15) \\\ \therefore \dfrac{{dT}}{{dt}} = -k(T - 15) \\\
where ‘k’ is any constant and the negative sign denotes the loss of heat to the surroundings because the initial temperature of coffee is more than the surrounding temperature.

dT(T−15)=−kdt\dfrac{{dT}}{{(T - 15)}} = -kdt
Apply integration on both sides
∫dT(T−15)=∫−kdt\int {\dfrac{{dT}}{{(T - 15)}}} = \int {-kdt}

⇒log⁡(T−15)=−kt+log⁡c \Rightarrow \log (T - 15) = -kt + \log c, where logc is the integration constant
⇒log⁡((T−15)c)=−kt\Rightarrow \log \left( {\dfrac{{(T - 15)}}{c}} \right) = -kt
((T−15)c)=e−kt⇒(T−15)=ce−kt\left( {\dfrac{{(T - 15)}}{c}} \right) = {e^{-kt}} \Rightarrow (T - 15) = c{e^{-kt}}
Given data: t=0, T=100
⇒100−15=ceo=c ⇒c=85 ⇒T−15=85e−kt→(1)  \Rightarrow 100 - 15 = c{e^o} = c \\\ \Rightarrow c = 85 \\\ \Rightarrow T - 15 = 85{e^{-kt}} \to (1) \\\
Now when t=5 minutes, T=60
⇒60−15=85e−5k ⇒e−5k=4585=917  \Rightarrow 60 - 15 = 85{e^{-5k}} \\\ \Rightarrow {e^{-5k}} = \dfrac{{45}}{{85}} = \dfrac{9}{{17}} \\\
Now we have to find out the temperature after a further interval of 5 minutes.
At t=10t = 10 minutes from equation (1)
⇒T−15=85e−kt ⇒T−15=85e−10k=85(917)2=23.82 ∴T=38.820C  \Rightarrow T - 15 = 85{e^{-kt}} \\\ \Rightarrow T - 15 = 85{e^{-10k}} = 85{\left( {\dfrac{9}{{17}}} \right)^2} = 23.82 \\\ \therefore T = {38.82^0}C \\\
Hence the temperature after 10 minutes is 38.820C{38.82^0}C.

Note: As you can see in these types of problems we have to use Newton's law of cooling. Based on the equation we got after integration, we can say that there is an exponential decrease in the temperature of the coffee wrt time.