If you place a cup of 210 degree fahrenheit coffee on a table in a room that is 68 degrees fareinheit, and ten minutes later, it is 200 degrees fareinheit. how long will it be before the coffee is 180 degrees fareinheit? use newtons law of cooling

Newton's Law of Cooling

Tf=Ts+(Ti-Ts)e^(-kt)   where Tf is temp at time t, Ts is temp of surroundings, Ti is temp of object/fluid.  So we need to find k first.

200=68+(210-68)e^(-10k)

132=142e^(-10k)

132/142=e^(-10k)

ln(132/142)=-10k

k=-ln(132/142)/10

k≈0.0073  so 

T(t)=68+142e^(-0.0073t)  so how long until it reaches 180°?

180=68+142e^(-0.0073t)

112=142e^(-0.0073t)

112/142=e^(-0.0073t)

ln(112/142)=-0.0073t

t= -ln(112/142)/(0.0073)

t≈32.51 minutes