Newtons law of cooling?!

when a hot object is placed in a water bath whose temperature is 30 degrees, it cools from 130 to 80 degrees in 100 seconds. In another bath, the cooling occurs in 200 seconds. find the temperature of the second bath.

Newton's Law of Cooling

T0 = Initial temperature -- 130
TA = Ambient temperature -- 30
t = time elapsed --100 seconds
T = Temperature after time t
T=TA+(T0-TA) e^kt
e^kt = (T-TA) / (T0-TA)
k = LN [ (T-TA) / (T0-TA)] / t

k = LN [ (80-30) / (130-30)] / 80
Constant of proportionality k = -0.006931471805599453
T = 30+100e^(-0.006931471805599453)(t)

Find cooled temperature after time 200
T = 30+100e^(-0.006931471805599453)(t)
Substitute t = 200
T = 30+100e^(-0.006931471805599453)(200)
Cooled temperature is 55 degrees