At 2 pm a hot coal was pulled out of a furnace and allowed t

At 2 p.m., a hot coal was pulled out of a furnace and allowed to cool at room temperature (75°F). If, after 10 minutes, the temperature of the coal was 415°F, and after 20 minutes, its temperature was 347°F, find the following.

Solution

Given

Tm=75

T(10)=415

T(20)=347

dT/dt= -k[T-T_m]

1/(T-75) dT= -k*dT

ln[T-75]=-Kt+c

T(t)=75+c_1*e^[-kt]

T(10)=415=75+c_1*e^[-10*t]

340=c_1*e^-10k.....(1)

As same T(20)

T(20)=347=75+c_1*e^[-20k]

272=c_1\" e^[-20k]...(2)

(1)/(2)

340/272=[c_1*e^-10k]/[c_1*e^-20k]

By sloving k=2.07

Then find the following ..which are not mentioned on the question