Given: $dT/dt=k(T-40)$Initial temperature of soda, $T_0=70°F$Temperature of soda after 1 hour in the refrigerator $=64°F$To find: At what time the soda's temperature be $47°F$?We need to find the time taken to cool from $64°F$ to $47°F$.Let's integrate both sides of the differential equation.$$\frac{dT}{dt}=k(T-40)$$
$$\frac{dT}{T-40}=kdt$$
Integrating both sides,
$$\int\frac{dT}{T-40}=\int kdt$$
On solving the above expression, we get
$$\ln|T-40|=kt+C$$
where $C$ is the constant of integration.Now, we can write this expression as:
$$|T-40|=e^{kt+C}$$
As $T-40>0$ for all $T>40°F$, we can remove the absolute value signs.
$$T-40=e^{kt+C}$$
$$T=40+Ce^{kt}$$To find the value of $C$, we can use the given information.Initial temperature, $T_0=70°F$When $t=0$,$T=70°F$Thus, $70=40+C$Or, $C=30$Using the value of $C$, we can write the expression for temperature as:$$T=40+30e^{kt}$$When $t=1$ hour, $T=64°F$.$$64=40+30e^{k(1)}$$
$$\implies 24=30e^k$$
$$\implies e^k=0.8$$
$$\implies k=-0.223$$
Thus, the equation for the temperature of the soda is given by:
$$T=40+30e^{-0.223t}$$To find the time required to cool the soda from $64°F$ to $47°F$:$T=47°F$We can write the above equation as:
$$47=40+30e^{-0.223t}$$
$$\implies 7=30e^{-0.223t}$$
$$\implies e^{-0.223t}=0.2333$$
Taking the natural log of both sides,
$$\ln e^{-0.223t}=\ln 0.2333$$
$$\implies -0.223t=\ln 0.2333$$
$$\implies t=\frac{\ln 0.2333}{-0.223} \approx 6.52 \ \text{hours (rounded to two decimal places)}$$Thus, the soda's temperature will be $47°F$ after approximately $6.52$ hours in the refrigerator.