A 200 F cup of coffee is placed in a 70 F room After 45 minu

A 200 F cup of coffee is placed in a 70 F room. After 45 minutes the temperature of the soup is 150 F. What will the temperature of the coffee be in another hour?

Solution

Given that

initial temperature of coffee To = 200 F

final temperature of coffee Tt = 150 F

room temperature Ta = 70 F

time = 45 min

( T(t) - Ta ) = ( To - Ta ) e^-kt

Solve for k:

T(t) - Ta = (To - Ta) e^-kt
150 - 70 = ( 200 - 70 )e^-k(45)

80 = 130 e^-k(45)

e^-k(45) = 80/130

-k.45 = In (80/130)

k.45 = In (130/80)

k = 0.0108

Calulation of temperature of coffee in another 1 hr:

   Initial To = 150 F

Final Tt = ?

Room temp Ta = 70 F

k = 0.0108

t = 1 hr = 60 min

T(t) - Ta = (To - Ta) e^-kt

T(t) - 70 = (150 -70)e^-(0.0108 x 60)

T(t) - 70 = 41.8

T(t) = 41.8 + 70

= 111.8 F

T(t) = 111.8 F

Therefore,

temperature of coffee in another 1 hr = 111.8 F