How long will it take for the can to cool off

I got a beverage in a can whose temperature is 21 degrees Celsius. To enjoy it cold at the perfect temperature of 9 degrees celsius, i set my refrigerator to 0 degrees celsius. The temperature T(t) of the can at time t measured in seconds satisfies Newton's Cooling Law equation:
dT/dt =-.0001T
How long will it take for the can to cool off to 9 degrees Celsius?

dT/dt = -.0001T

dT/T = -.0001dt

(integrate both sides)
ln(T) = -.0001t + c

T= Ce^(-.0001)t

when t = 0, T = 21

so...

21 = Ce^(-.0001 * 0) = Ce^(0) = C

so, T = 21e^(-.0001t)

now to get to 9 degrees, solve for when T = 9

9 = 21e^(-.0001t)

(9/21) = e^(-.0001t)
ln(9/21) = -.0001t

t = [ln(9/21)/(-.0001)] = 8473 seconds = 141.2 min = 2.35hours