When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round your answers to two decimal places.)
(a) What is the temperature of the drink after 40 minutes?
°C
(b) When will its temperature be 15°C?
min
My work is here: I not sure I am doing it right. 
Best Answer
Notice, we have $$T=Ce^{kt}+T_s$$ Setting $T=T_0$ (initial temperature of cold drink) at $t=0$ then we have $$T_0=Ce^{0}+T_s\iff C=T_0-T_s$$
Now, we have $$T=(T_0-T_s)e^{kt}+T_s$$ $$e^{kt}=\frac{T-T_s}{T_0-T_s}$$ $$t=\frac{1}{k}\ln\left(\frac{T-T_s}{T_0-T_s}\right)\tag 1$$
Given condition: Temperature rise of cold drink from $\color{red}{T_0=5^\circ \ C\to T=10^\circ \ C}$ in time $t=25\ minutes$ & $\color{red}{T_s=20^\circ \ C}$ substituting values in (1) we get $$25=\frac{1}{k}\ln\left(\frac{10-20}{5-20}\right)$$ $$k=\frac{1}{25}\ln\left(\frac{2}{3}\right)$$
a) Temperature $\color{red}{T}$ after $t=40 \ minutes$, setting corresponding values in (1), we get $$40=\frac{1}{\frac{1}{25}\ln\left(\frac{2}{3}\right)}\ln\left(\frac{T-20}{5-20}\right)$$ $$\ln\left(\frac{20-T}{15}\right)=\frac{40}{25}\ln\left(\frac{2}{3}\right)$$ $$\frac{20-T}{15}=\left(\frac{2}{3}\right)^{8/5}$$ $$\color{red}{T}=20-15\left(\frac{2}{3}\right)^{8/5}\approx \color{blue}{12.15^\circ\ C}$$
b) Time $\color{red}{t}$ when the temperature is $\color{red}{T=15^\circ C}$, setting corresponding values in (1), we get $$t=\frac{1}{\frac{1}{25}\ln\left(\frac{2}{3}\right)}\ln\left(\frac{15-20}{5-20}\right)$$ $$=\frac{25 \ln\left(\frac{1}{3}\right)}{\ln\left(\frac{2}{3}\right)}$$
$$\color{red}{t}\approx \color{blue}{67.74\ minutes}$$