Answer:
the temperature of the object after one hour is 48.1 degree F
Explanation:
Given data
room temp = 65 F
outside temp = 35 F
5 min temperature = 63 F
to find out
the temperature of the object after one hour
solution
according to Newton's Law
dT/dt = -K( T - 35)
dT/T-35 = -k dt
integrate both side
ln (T-35 ) = -kt +c
we take here T(0) = 65
ln (65 - 35) = -kt +c
so c = ln(30)
so it will be
ln (T-35 ) = -kt + ln30 .....................1
now we again take T(5) = 63 according to question
ln (63-35 ) = -kt + ln30
ln 28 = -5k + ln 30
so 5k = ln (30/28)
k = 1/5 ln(15/14)
now put this k value in equation 1
ln (T-35 ) = -1/5 ln(15/14)t + ln30
T-35 = 30 [tex](15/14)^{-t/5}[/tex]
so after 1 hour temperature will be
T(60) = 30 [tex](15/14)^{-t/5}[/tex] +30
T(60) = 30 [tex](15/14)^{-60/5}[/tex] +30
T(60) = 35 +30 [tex](15/14)^{-12}[/tex]
T(60) = 48.1 degree F
the temperature of the object after one hour is 48.1 degree F