Please help me out with differential equations

Question

When a cake is removed from an oven, its temperature is measured at 300° F. Three

minutes later its temperature is 200° F. How long will it take for the cake to cool off to a

room temperature of 70° F?

Answer 1

Time taken by the cake to cool from 300 to 200 , let it be t = 3 min = 3*60 = 180 sec

Now rate of cooling = (300 - 200)/180 F/sec = 100/180 F/sec = 0.556 F/sec = rate

Fall in temperature = 200-70 = 130F

Now time taken for cake to cool = (fall in temperature)/ rate

Thus time taken for it to cool down = 130/0.556 sec = 72.68s = 1.204 min