newton's law of cooling ( Differential equations)

Question

A cold drink is brought into a warm room with initial temperature at 40F . A few mins later the drink is found to be 46F , after the same length of time , it becomes 51F . Use newton law of cooling to determine the temperature of the room ?

T(0)=40 dT/dt = k(T-Tm) , K is constant smaller than 0 T(t) = Tm + (T0 - Tm)e^kt T=Temperature , t=Time , Tm = Temperature of the room We need to find Tm with this data that are shown in the question

Answer 1

HINT

We have that

$$\frac{dT}{dt}=r\Delta T\implies T(t)=T_{room}+(T_0-T_{room})e^{-rt}$$

then write down the givens and use $\log$ to find out $T_{room}$.

• $46=T_{room}+(40-T_{room})e^{-rt}\implies e^{-rt}=\frac{46-T_{room}}{40-T_{room}}\implies -rt=\log\left(\frac{46-T_{room}}{40-T_{room}}\right)$
• $51=T_{room}+(40-T_{room})e^{-2rt}\implies e^{-2rt}=\frac{51-T_{room}}{40-T_{room}}\implies -2rt=\log\left(\frac{51-T_{room}}{40-T_{room}}\right)$

thus

$$2\log\left(\frac{46-T_{room}}{40-T_{room}}\right)=\log\left(\frac{51-T_{room}}{40-T_{room}}\right)\iff \left(\frac{46-T_{room}}{40-T_{room}}\right)^2=\left(\frac{51-T_{room}}{40-T_{room}}\right)$$

$$(46-T_{room})^2=(51-T_{room})(40-T_{room})\implies T_{room}=76 $$