Question

An object is placed in a 68 degree room. write a differential equation for H, the temperature of the object at time t.
dH/dt=k(68-H)

Give the general solution for the differential equation.
H=68-Ae^(-kt)

The temperature of the object is 40 degrees initially, and 48 degrees one hour later. Find teh temperature of the object after 3 hours.

I used 48=68-40e^(-k*1) to find k which I found to be about .693. I'm not sure if I did this correctly.

Then I used that k in the equation
H=68-40e^(-.693*t)=63 degrees
but the answer in the back of my book says 57.8 degrees. Help?